3 Mark Question

Question 13 Marks

Find the cube root of each of the following natural numbers:
157464

Answer

By prime factorization method,
$=3_\sqrt{157464}$
$=3_\sqrt{2\times2\times2\times3\times3\times3\times3\times3\times3\times3\times3\times3}$
$=3_\sqrt{2^3\times3^3\times3^3\times3^3}$
= 2 × 3 × 3 × 3
= 54

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Question 23 Marks

Find the cube roots of the following integers:
$-32768$

Answer

We have,
$=\sqrt[3]{-32768}$
$=-\sqrt[3]{32768}$
To find the cube root of 32768, we use the method of unit digits.
Let us consider the number 32768.
The unit digit is 8; therefore, the unit digit in the cube root of 32768 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 32.
Now, 3 is the largest number whose cube is less than or equal to $32\left(3^3<3^2<4^3\right)$.
Therefore, the tens digit of the cube root 32768 is 3.
$\therefore\sqrt[3]{32768}=32$
$\Rightarrow \sqrt[3]{-32768}$
$= -\sqrt[3]{32768}$
$=-32$

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Question 33 Marks

Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that.
20346417 = 9261 × 2197

Answer

To find the cube root, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for two integers a and b.
Now
$=\sqrt[3]{20346417}$
$=\sqrt[3]{9261\times2197}$
$=\sqrt[3]{9261}\times\sqrt[3]{2197}$ (By the above property)
$=\sqrt[3]{3\times3\times3\times7\times7\times7}\times\sqrt[3]{13\times13\times13}$ (By prime factorisation)
$=\sqrt[3]{\{3\times3\times3\}\times\{7\times7\times7\}}\times\sqrt[3]{\{13\times13\times13\}}$(By prime factorisation)
$=3\times7\times13$
$=273$
Thus, the answer is 273.

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Question 43 Marks

For each of the non-perfect cubes in Q. No. 20 find the smallest number by which it must beMultiplied so that the product is a perfect cube.

Answer

The only non-perfect cube in question number 20 is 243.
On factorising 243 into prime factors, we get:
243 = 3 × 3 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
243 = {3 × 3 × 3} × 3 × 3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is multiplied by 3, the factors can be grouped into triples of equal factors such that no factor is left over.

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Question 53 Marks

Show that:
$\frac{\sqrt[3]{-512}}{\sqrt[3]{343}}=\sqrt[3]{\frac{-512}{343}}$

Answer

$\text{LHS}=\frac{\sqrt[3]{-512}}{\sqrt[3]{343}}$
$=\frac{-\sqrt[3]{512}}{\sqrt[3]{343}}$
$=\frac{\sqrt[3]{\{2\times2\times2\} \times\{2\times2\times2\}\times\{2\times2\times2\}}}{\sqrt[3]{7\times7\times7}}$
$=\frac{-(2\times2\times2)}{7}$
$=\frac{-8}{7}$
$\text{RHS}=\sqrt[3]\frac{{-512}}{{343}}$
$=\sqrt[3]\frac{{(-2)\times(-2)\times(-2)\times(-2)\times(-2)\times(-2)\times(-2)\times(-2)\times(-2)}}{{7\times7\times7}}$
$=\sqrt[3]{\frac{(-2)\times(-2)\times(-2)}{7}\times\frac{(-2)\times(-2)\times(-2)}{7}\times\frac{(-2)\times(-2)\times(-2)}{7}}$
$=\sqrt[3]{(\frac{-8}{7})^3}$
$=\frac{-8}{7}$
Because LHS is equal to RHS, the equation is true.

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Question 63 Marks

Find the side of a cube whose volume is $\frac{24389}{216}\text{m}^3.$

Answer

Volume of a cube with side s is given by:
$\text{V}=\text{s}^3$
$\therefore\text{s}=\sqrt[3]{\text{V}}$
$=\sqrt[3]{\frac{24389}{216}}$
$=\frac{\sqrt[3]{24389}}{\sqrt[3]{216}}$
$=\frac{\sqrt[3]{29\times29\times29}}{\sqrt[3]{2\times2\times2\times3\times3\times3}}$ (By prime factorisation)
$=\frac{29}{2\times3}$
$=\frac{29}{6}$
Thus, the length of the side is $=\frac{29}{6}\text{m.}$

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Question 73 Marks

Find the cube root of each of the following natural numbers:
4913

Answer

By prime factorization method,
$=3_\sqrt{4913}$
$=3_\sqrt{17\times17\times17}$
$=17$

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Question 83 Marks

Which of the following numbers are cubes of negative integers.
$-64$

Answer

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer m, $m ^3$ is the cube of - m .
On factorising 64 into prime factors, we get:
$64=2 \times 2 \times 2 \times 2 \times 2 \times 2$
On grouping the factors in triples of equal factors, we get:
$64=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\}$
It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube. This implies that -64 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get:
$2 \times 2=4$
This implies that 64 is a cube of 4 .
Thus, -64 is the cube of -4 .

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Question 93 Marks

Evaluate:
$\sqrt[3]{96}\times\sqrt[3]{144}$

Answer

96 and 122 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for any two integers a and b.
$\therefore\sqrt[3]{96}\times\sqrt[3]{144}$
$=\sqrt[3]{{96}\times{144}}$
$=\sqrt[3]{(2\times2\times2\times2\times 2\times3)\times(2\times2\times2\times2\times2\times3)}$ (By prime factorisation)
$=\sqrt[3]{\{2\times2\times2\}\times\{2\times2\times2\}\times\{2\times2\times2\}\times\{3\times3\times3\}}$
$=2\times2\times2\times3$
$=24$
Thus, the answer is 24.

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Question 103 Marks

Evaluate the following:
$\Big\{(5^2+12^2)^{\frac{1}{2}}\Big\}^3$

Answer

To evaluate the value of the given expression, we can proceed as follows:
$\Big\{(5^2+12^2)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(25+144)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(169)^{\frac{1}{2}}\Big\}^3$
$=\Big\{\sqrt{(169)}\Big\}^3$
$=\Big\{\sqrt{(13\times13)}\Big\}^3$
$=\{13\}^3$
$13\times13\times13=2197$

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Question 113 Marks

By taking three different values of $n$ verify the truth of the following statement:If a natural number n is of the form $3 p+2$ then $n^3$ also a number of the same type.

Answer

Three natural numbers of the form $(3 p+2)$ can be written by choosing $p=1,2,3 \ldots$ etc.
Let three such numbers be 5,8 and 11 .
Cubes of the three chosen numbers are:
$5^3=125,8^3=512 \text { and } 11^3=1331$
Cubes of 5, 8, and 11 can be expressed as:
$125=3 \times 41+2$, which is of the form $(3 p+2)$ for $p=41$
$512=3 \times 170+2$, which is of the form $(3 p+2)$ for $p=170$
$1331=3 \times 443+2$, which is of the form $(3 p+2)$ for $p=443$
Cubes of 5,8 , and 11 could be expressed as the natural numbers of the form $(3 p+2)$ for some natural number $p$. Hence, the statement is verified.

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Question 123 Marks

Making use of the cube root table, find the cube root 1100

Answer

We have:
1100 = 11 × 100
$\therefore\sqrt[3]{1100}$
$=\sqrt[3]{11\times100}$
$=\sqrt[3]{11}\times\sqrt[3]{100}$
By the cube root table, we have:
$=\sqrt[3]{11}=2.224$ and $\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{1100}$
$=\sqrt[3]{11}\times\sqrt[3]{100}$
$=2.224\times4.642$
$=10.323$ (Up to three decimal places)
Thus, the answer is 10.323.

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Question 133 Marks

Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that.
3048625 = 3375 × 729

Answer

To find the cube root, we use the following property: $\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for two integers a and b. Now $=\sqrt[3]{3048625}$ $=\sqrt[3]{3375\times729}$ $=\sqrt[3]{3375}\times\sqrt[3]{729}$ (By the above property) $=\sqrt[3]{3\times3\times3\times5\times5\times5}\times\sqrt[3]{9\times9\times9}$ (By prime factorisation) $=\sqrt[3]{\{3\times3\times3\}\times\{5\times5\times5\}}\times\sqrt[3]{\{9\times9\times9\}}$ $=3\times5\times9$ $=135$ Thus, the answer is 135.

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Question 143 Marks

Evaluate:
$\sqrt[3]{121}\times\sqrt[3]{297}$

Answer

121 and 297 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for any two integers a and b.
$\therefore\sqrt[3]{121}\times\sqrt[3]{297}$
$=\sqrt[3]{{121}\times{297}}$
$=\sqrt[3]{(11\times11\times11)\times (3\times3\times3)}$ (By prime factorisation)
$=\sqrt[3]{\{2\times2\times2\}\times\{3\times3\times3\}\times\{5\times5\times5\}}$
$=11\times3$
$=33$
Thus, the answer is 33.

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Question 153 Marks

Observe the following pattern:
$1^3=1$
$1^3+2^3=(1+2)^2$
$1^3+2^3+3^3=(1+2+3)^2$
Write the next three rows and calculate the value of $1^3+2^3+3^3+\ldots+9^3+10^3$ by the above pattern.

Answer

$1^3=1$
$1^3+2^3=(1+2)^2$
$1^3+2^3+3^3=(1+2+3)^2$
$1^3+2^3+3^3+4^3=(1+2+3+4)^2$
$1^3+2^3+3^3+4^3+5^3=(1+2+3+4+5)^2$
$1^3+2^3+3^3+4^3+5^3+6^3=(1+2+3+4+5+6)^2$
Now, from the ad
$1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3+10^3=(1+2+3+4+5+6+7+8+9+10)^2$
$55^2=3025$
Thus, the required value is 3025 .

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Question 163 Marks

Three numbers are in the ratio $1: 2: 3$. The sum of their cubes is 98784 . Find the numbers.

Answer

Let the numbers are $= x , 2 x$ and 3 x
According to the question,
$x^3+(2 x)^3+(3 x)^3=98784$
$x^3+8 x^3+27 x^3=98784$
$36 x^3=98784$
$x^3=\frac{98784}{36}=2744$
$x=3_{\sqrt{2744}}$
$=3_{\sqrt{2 \times 2 \times 2 \times 7 \times 7 \times 7}}$
$=2 \times 7=14$
So, the numbers are,
$x=14$
$2 x=2 \times 14=28$
$3 x=3 \times 14=42$

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Question 173 Marks

Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer.
$-5832$

Answer

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer m , $m ^3$ is the cube of - m .
On factorising 5832 into prime factors, we get:
$5832=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors, we get:
$5832=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{3 \times 3 \times 3\}$
It is evident that the prime factors of 5832 can be grouped into triples of equal factors and no factor is left over. Therefore, 5832 is a perfect cube. This implies that -5832 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get:
$2 \times 3 \times 3=18$
This implies that 5832 is a cube of 18 .
Thus, -5832 is the cube of -18 .

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Question 183 Marks

Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that.
210644875 = 42875 × 4913

Answer

To find the cube root, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for two integers a and b.
Now
$=\sqrt[3]{210644875}$
$=\sqrt[3]{42875\times4913}$
$=\sqrt[3]{42875}\times\sqrt[3]{4913}$ (By the above property)
$=\sqrt[3]{5\times5\times5\times7\times7\times7}\times\sqrt[3]{17\times17\times17}$ (By prime factorisation)
$=\sqrt[3]{\{5\times5\times5\}\times\{7\times7\times7\}}\times\sqrt[3]{\{17\times17\times17\}}$(By prime factorisation)
$=3\times7\times17$
$=595$
Thus, the answer is 595.

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Question 193 Marks

Evaluate:
$\sqrt[3]{36}\times\sqrt[3]{384}$

Answer

36 and 384 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for any two integers a and b.
$\therefore\sqrt[3]{36}\times\sqrt[3]{384}$
$=\sqrt[3]{{36}\times{384}}$
$=\sqrt[3]{(2\times2\times3\times3)\times(2\times2\times2\times2\times2\times2\times2\times3)}$ (By prime factorisation)
$=\sqrt[3]{\{2\times2\times2\}\times\{2\times2\times2\}\times\{2\times2\times2\}\times\{3\times3\times3\}}$
$=2\times2\times2\times3$
$=24$
Thus, the answer is 24.

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Question 203 Marks

Find The cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that.
57066625 = 166375 × 343

Answer

To find the cube root, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for two integers a and b.
Now
$=\sqrt[3]{57066625}$
$=\sqrt[3]{166375\times343}$
$=\sqrt[3]{166375}\times\sqrt[3]{343}$ (By the above property)
$=\sqrt[3]{5\times5\times5\times11\times11\times11}\times\sqrt[3]{7\times7\times7}$ (By prime factorisation)
$=\sqrt[3]{\{5\times5\times5\}\times\{11\times11\times11\}}\times\sqrt[3]{\{7\times7\times7\}}$(By prime factorisation)
$=3\times11\times7$
$=385$
Thus, the answer is 385.

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Question 213 Marks

Find the cube root of each of the following natural numbers:35937

Answer

By prime factorization method,
$=3_\sqrt{35937}$
$=3_\sqrt{3\times3\times3\times11\times11\times11}$
$=3_\sqrt{3^3\times11^3}$
$=3\times11$
$=33$

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Question 223 Marks

By taking three different values of $n$ verify the truth of the following statement:
If n leaves remainder 1 when divided by 3 , then $n ^3$ also leaves 1 as remainder when divided by 3 .

Answer

Three natural numbers of the form $(3 n+1)$ can be written by choosing $n =1,2,3 \ldots$ etc.
Let three such numbers be 4,7 and 10 .
Cubes of the three chosen numbers are:
$4^3=64,7^3=343 \text { and } 10^3=1000$
Cubes of 4, 7 and 10 can expressed as:
$64=3 \times 21+1$, which is of the form $(3 n+1)$ for $n=21$
$343=3 \times 114+1$, which is of the form $(3 n+1)$ for $n=114$
$1000=3 \times 333+1$, which is of the form $(3 n+1)$ for $n=333$
Cubes of 4,7 , and 104,7 , and 10 can be expressed as the natural numbers of the form $(3 n+1)$ for some natural number $n$. Hence, the statement is verified.

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Question 233 Marks

Evaluate the following:
$\sqrt[3]{\text{a}24}+\sqrt[3]{0.008}+\sqrt[3]{0.0064}$

Answer

To evaluate the value of the given expression, we need to proceed as follows:
$\sqrt[3]{\text{a}24}+\sqrt[3]{0.008}+\sqrt[3]{0.0064}$
$=\sqrt[3]{3\times3\times3}+\sqrt[3]{\frac{8}{1000}}+\sqrt[3]{\frac{64}{1000}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}}+{\frac{\sqrt[3]{64}}{\sqrt[3]{1000}}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{2\times2\times2}}{\sqrt[3]{1000}}}+{\frac{\sqrt[3]{4\times4\times4}}{\sqrt[3]{1000}}}$
$=3+\frac{2}{10}+\frac{4}{10}$
$=3+0.2+0.4$
$=3.6$
Thus, the answer is 3.6.

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Question 243 Marks

Find the cube root of each of the following natural numbers:
48228544

Answer

By prime factorization method,
$=3_\sqrt{48228544}$
$=3_\sqrt{2\times2\times2\times2\times2\times2\times7\times7\times7\times13\times13\times13}$
$=3_\sqrt{2^3\times2^3\times7^3\times13^3}$
= 2 × 2 × 7 × 13
= 364

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Question 253 Marks

Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer.
$-2744000$

Answer

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer m , - $m ^3$ is the cube of - m .
On factorising 2744000 into prime factors, we get:
$2744000=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7$
On grouping the factors in triples of equal factors, we get:
$2744000=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{5 \times 5 \times 5\} \times\{7 \times 7 \times 7\}$
It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744000 is a perfect cube. This implies that -2744000 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get:
$2 \times 2 \times 5 \times 7=140$
This implies that 2744000 is a cube of 140.

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Question 263 Marks

Which of the following numbers are cubes of negative integers.
$-2197$

Answer

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer $m$, $- m ^3$ is the cube of - m .
On factorising 2197 into prime factors, we get:
$2197=13 \times 13 \times 13$
On grouping the factors in triples of equal factors, we get:
$2197=\{13 \times 13 \times 13\}$
It is evident that the prime factors of 2197 can be grouped into triples of equal factors and no factor is left over. Therefore, 2197 is a perfect cube. This implies that -2197 is also a perfect cube
Now, collect one factor from each triplet and multiply, we get 13.
This implies that 2197 is a cube of 13.
Thus, -2197 is the cube of -13 .

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Question 273 Marks

Evaluate:
$125\sqrt[3]{\text{a}^6}-\sqrt[3]{125\text{a}^6}$

Answer

Property:
For any two integers a and $b, \sqrt[2]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$,
From the above property, we have:
$125 \sqrt[3]{a^6}-\sqrt[3]{125 a^6}$
$=125 \sqrt[3]{a^6}\left(\sqrt[3]{125} \times \sqrt[3]{a^6}\right)$
$=125 \times a^2-\left(5 \times a^2\right)$
$\left(\because \sqrt[3]{a \times a \times a\} \times\{a \times a \times a\}}=a \times a=a^2 and \sqrt[3]{125}=\sqrt{5 \times 5 \times 5}=5\right)$
$=125 a^2-5 a^2$
$=120 a^2$

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Question 283 Marks

Find the cube root of each of the following natural numbers:1728

Answer

By prime factorization method,
$=3_\sqrt{1728}$
$=3_\sqrt{2\times2\times2\times2\times2\times2\times3\times3\times3}$
$=3_\sqrt{2^3\times2^3\times3^3}$
$=2\times2\times3$
$=12$

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Question 293 Marks

Making use of the cube root table, find the cube root 5112

Answer

By prime factorisation, we have:
$5112=2^3 \times 3^2 \times 71$
$\Rightarrow \sqrt[3]{5112}$
$=2 \times \sqrt[3]{9} \times \sqrt[3]{71}$
By the cube root table, we have:
$\sqrt[3]{9}=2.080 \text { and } \sqrt[3]{71}=4.141$
$\therefore \sqrt[3]{5112}$
$=2 \times \sqrt[3]{9} \times \sqrt[3]{71}$
$=2 \times 2.080 \times 4.141$
$=17.227 \text { (upto three decimal places) }$
Thus, the required cube root is 17.227 .

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Question 303 Marks

Find the cube root of the following numbers:
-27 × 2744

Answer

Property:
For any two integers a and b, $\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}},$
From the above property, we have:
$\sqrt[3]{-27\times2744}$
$=\sqrt[3]{-27}\times\sqrt[3]{2744}$
$-=\sqrt[3]{1728}\times\sqrt[3]{216}$ (For any positive integer $\text{x},\sqrt[3]{-\text{x}}=-\sqrt[3]{\text{x}}$)
Cube root using units digit:
Let us consider the number 2744.
The unit digit is 4; therefore, the unit digit in the cube root of 2744 will be 4.
After striking out the units, tens, and hundreds digits of the given number, we are left with 2.
Now, 1 is the largest number whose cube is less than or equal to 2.
Therefore, the tens digit of the cube root of 2744 is 1
$\therefore\sqrt[3]{2744}=14$
Thus
$\sqrt[3]{-27\times2744}$
$=-\sqrt[3]{27}\times\sqrt[3]{2744}$
$=-3\times6=-42$

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Question 313 Marks

Find the cube root of each of the following natural numbers:
33698267

Answer

By prime factorization method,
$=3_\sqrt{33698267}$
$=3_\sqrt{17\times17\times17\times19\times19\times19}$
$=3_\sqrt{17^3 \times19^3}$
= 17 × 19
= 323

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Question 323 Marks

Which of the following numbers are cubes of negative integers.
$-2744$

Answer

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer $m$, $m ^3$ is the cube of - m .
On factorising 2744 into prime factors, we get:
$2744=2 \times 2 \times 2 \times 7 \times 7 \times 7$
On grouping the factors in triples of equal factors, we get:
$2744=\{2 \times 2 \times 2\} \times\{7 \times 7 \times 7\}$
It is evident that the prime factors of 2744 can be grouped into triples of equal factors and no factor is left over. Therefore, 2744 is a perfect cube. This implies that -2744 is also a perfect cube.
Now, collect one factor from each triplet and multiply, we get:
$2 \times 7=14$
This implies that 2744 is a cube of 14 .
Thus, -2744 is the cube of -14 .

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Question 333 Marks

Making use of the cube root table, find the cube root 7800

Answer

We have:
7800 = 78 × 100
$\therefore\sqrt[3]{7800}$
$=\sqrt[3]{78\times100}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$
By the cube root table, we have:
$=\sqrt[3]{78}=4.273$ and $\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{7800}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$
$=4.273\times4.642$
$=19.835$ (Up to three decimal places)
Thus, the answer is 19.835.

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Question 343 Marks

Evaluate:
$\sqrt[3]{100}\times\sqrt[3]{270}$

Answer

100 and 270 are not perfect cubes; therefore, we use the following property:
$\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ for any two integers a and b.
$\therefore\sqrt[3]{100}\times\sqrt[3]{270}$
$=\sqrt[3]{{100}\times{270}}$
$=\sqrt[3]{(2\times2\times5\times5)\times (2\times3\times3\times3\times5)}$ (By prime factorisation)
$=\sqrt[3]{\{2\times2\times2\}\times\{3\times3\times3\}\times\{5\times5\times5\}}$
$=2\times3\times5$
$=30$
Thus, the answer is 30.

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Question 353 Marks

Find the cube root of each of the following natural numbers:
1157625

Answer

By prime factorization method,
$=3_\sqrt{1157625}$
$=3_\sqrt{3\times3\times3\times5\times5\times5\times7\times7\times7}$
$=3_\sqrt{3^3\times5^3\times7^3}$
= 2 × 5 × 7
= 105

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Question 363 Marks

Find the cube root of each of the following natural numbers:
17576

Answer

By prime factorization method,
$=3_\sqrt{17576}$
$=3_\sqrt{2\times2\times2\times13\times13\times13}$
$=3_\sqrt{2^3\times13^3}$
$=2\times13$
$=26$

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Question 373 Marks

Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824

Answer

We know that the cubes of all even natural numbers are even.
The numbers 216, 512, 1000 and 13824 are cubes of even natural numbers.
The numbers 216, 512, 1000 and 13824 are even and it could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.
Thus, the cubes of even natural numbers are 216, 512, 1000 and 13824.

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Question 383 Marks

Evaluate the following:
$\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$

Answer

To evaluate the value of the given expression, we need to proceed as follows:
$\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$
$=\sqrt[3]{10\times10\times10}+\sqrt[3]{\frac{8}{1000}}-\sqrt[3]{\frac{125}{1000}}$
$=\sqrt[3]{10\times10\times10}+{\frac{\sqrt[3]{8}}{\sqrt[3]{1000}}}-{\frac{\sqrt[3]{125}}{\sqrt[3]{1000}}}$
$=\sqrt[3]{3\times3\times3}+{\frac{\sqrt[3]{2^3}}{\sqrt[3]{1000}}}-{\frac{\sqrt[3]{5^3}}{\sqrt[3]{1000}}}$
$=10+\frac{2}{10}-\frac{5}{10}$
$=10+0.2-0.5$
$=9.7$
Thus, the answer is 9.7.

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Question 393 Marks

For each of the non-perfect cubes in Q. No. 20 find the smallest number by which it must be.
Divided so that the quotient is a perfect cube.

Answer

Thus, 243 should be multiplied by 3 to make it a perfect cube.On factorising 243 into prime factors, we get:
243 = 3 × 3 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
243 = {3 × 3 × 3} × 3 × 3
It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is divided by (3 × 3 = 9), the factors can be grouped into triples of equal factors such that no factor is left over.

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Question 403 Marks

Which of the following numbers are cubes of negative integers.
$-1056$

Answer

In order to check if a negative number is a perfect cube, first check if the corresponding positive integer is a perfect cube. Also, for any positive integer m , $- m ^3$ is the cube of - m .
On factorising 1056 into prime factors, we get:
$1056=2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 11$
On grouping the factors in triples of equal factors, we get:
$1056=\{2 \times 2 \times 2\} \times 2 \times 2 \times 3 \times 11$
It is evident that the prime factors of 1056 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 1056 is not a perfect cube. This implies that -1056 is not a perfect cube as well.

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Question 413 Marks

Evaluate the following:
$\Big\{(6^2+8^2)^{\frac{1}{2}}\Big\}^3$

Answer

To evaluate the value of the given expression, we can proceed as follows:
$\Big\{(6^2+8^2)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(36+64)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(100)^{\frac{1}{2}}\Big\}^3$
$=\Big\{\sqrt{(100)}\Big\}^3$
$=\Big\{\sqrt{(10\times10)}\Big\}^3$
$=\{10\}^3$
$10\times10\times10=1000$

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Question 423 Marks

Find the cube roots of the following integers:
-2744000

Answer

We have,
Cube root of -125
$=\sqrt[3]{-2744000}$
$=-\sqrt[3]{2744000}$
To find the cube root of 2744000, we use the method of factorisation.
2 × 2 × 5 × 7 = 140
On factorising 2744000 into prime factors, we get:
2744000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7
On grouping the factors in triples of equal factors, we get:
2744000 = {2 × 2 × 2} × {2 × 2 × 2} × {5 × 5 × 5} × {7 × 7 × 7}
It is evident that the prime factors of 2744000 can be grouped into triples of equal factors and no factor is left over.
Now, collect one factor from each triplet and multiply; we get:
2 × 2 × 5 × 7 =140
This implies that 2744000 is a cube of 140.
Hence,
$=\sqrt[3]{-2744000}$
$= -\sqrt[3]{2744000}$
$=-140$

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Question 433 Marks

Making use of the cube root table, find the cube root 7000

Answer

We have:
700 = 70 × 100
$\therefore\sqrt[3]{7000}$
$=\sqrt[3]{7\times1000}$
$=\sqrt[3]{7}\times\sqrt[3]{1000}$
By the cube root table, we have:
$=\sqrt[3]{7}=1.913$ and $\sqrt[3]{1000}=10$
$\therefore\sqrt[3]{7000}$
$=\sqrt[3]{7}\times\sqrt[3]{1000}$
$=1.913\times10$
$=19.13$

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Question 443 Marks

Find the cube root of each of the following natural numbers:
2744

Answer

By prime factorization method,
$=3_\sqrt{2774}$
$=3_\sqrt{2\times2\times2\times7\times7\times7}$
$=3_\sqrt{2^3\times7^3}$
$=2\times7$
$=14$

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Question 453 Marks

Making use of the cube root table, find the cube root 9800

Answer

We have:
9800 = 98 × 100
$\therefore \sqrt[3]{9800}$
$=\sqrt[3]{98\times100}$
$=\sqrt[3]{98}\times\sqrt[3]{100}$
By cube root table, we have:
$\sqrt[3]{98}=4.610 $ and $\sqrt[3]{100}=4.642 $
$\therefore\sqrt[3]{9800}$
$=\sqrt[3]{98}\times\sqrt[3]{100}$
$= 4.610 \times4.642=21.40$ (upto three decimal places)
Thus, the required cube root is 21.40.

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Question 463 Marks

Find the cube root of the following rational numbers:
$\frac{686}{-3456}$

Answer

Let us consider the following rational number:
$\frac{686}{-3456}$
Now
$\sqrt[3]{\frac{686}{-3456}}$
$\sqrt[3]{\frac{2\times7^3}{2^7\times3^3}}$ 686 and 3456 are not perfect cubes; therefore, we simplify it as $\frac{686}{3456}$ by prime factorisation.)
$=-\sqrt[3]{\frac{7^3}{2^6\times3^3}}$
$=\frac{-\sqrt[3]{7^3}}{\sqrt[3]{2^6\times^3}}$
$=\frac{{-7}}{\sqrt[3]{2^3\times2^3\times3^3}}$
$=\frac{{-7}}{\sqrt[3]{2\times2\times3}}$
$=\frac{-7}{12}$
$\Big(\because\sqrt[3]{\frac{\text{a}}{\text{b}}}=\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}\Big)$

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Question 473 Marks

Find the cube root of each of the following natural numbers:
134217728

Answer

By prime factorization method,
$=3_\sqrt{134217728}$
$=3_\sqrt{2^{27}}$
$=2^9$
$=512$

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Question 483 Marks

Find the cube roots of the following integers:
$-733571$

Answer

We have,
$=\sqrt[3]{-753571}$
$=-\sqrt[3]{753571}$
To find the cube root of 753571, we use the method of unit digits.
Let us consider the number 753571 .
The unit digit is 1 ; therefore the unit digit in the cube root of 753571 will be 1.
After striking out the units, tens and hundreds digits of the given number, we are left with 753.
Now, 9 is the largest number whose cube is less than or equal to $753\left(9^3<753<10^3\right)$.
Therefore, the tens digit of the cube root 753571 is 9 .
$\therefore \sqrt[3]{753571}=91$
$=\sqrt[3]{-753571}$
$=-\sqrt[3]{753571}$
$=-91$

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