Question 13 Marks
Find the cube root of each of the following natural numbers: $157464$
AnswerBy prime factorization method, $=3_\sqrt{157464}$
$=3_\sqrt{2\times2\times2\times3\times3\times3\times3\times3\times3\times3\times3\times3}$
$=3_\sqrt{2^3\times3^3\times3^3\times3^3} = 2 \times 3 \times 3 \times 3 = 54$
View full question & answer→Question 23 Marks
Find the cube roots of the following integers: $-32768$
AnswerWe 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 $32\ (3^3 < 3^2 < 4^3).$
Therefore, the tens digit of the cube root $32768$ is $3.$
$\therefore\sqrt[3]{32768}=32$
$\Rightarrow \sqrt[3]{-32768}$
$= -\sqrt[3]{32768}$
$=-32$
View full question & answer→Question 33 Marks
Find The cube roots of the numbers $3048625, 20346417, 210644875, 57066625$ using the fact that. $20346417 = 9261 \times 2197$
AnswerTo 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 $273.$
View full question & answer→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.
AnswerThe only non-perfect cube in question number $20$ is $243.$
On factorising $243$ into prime factors,
we get: $243 = 3 \times 3 \times 3 \times 3 \times 3$ On grouping the factors in triples of equal factors,
we get: $243 = {3 \times 3 \times 3} \times 3 \times 3$ 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.
View full question & answer→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.
View full question & answer→Question 63 Marks
Find the side of a cube whose volume is $\frac{24389}{216}\text{m}^3.$
AnswerVolume 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 $=\frac{29}{6}\text{m.}$
View full question & answer→Question 73 Marks
Find the cube root of each of the following natural numbers:
$4913$
AnswerBy prime factorization method,
$=3_\sqrt{4913}$
$=3_\sqrt{17\times17\times17}$
$=17$
View full question & answer→Question 83 Marks
Which of the following numbers are cubes of negative integers. $−64$
AnswerIn 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:
$64 = 2 × 2 × 2 × 2 × 2 × 2$
On grouping the factors in triples of equal factors, we get:
$64 = {2 × 2 × 2} × {2 × 2 × 2}$
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:
$2 × 2 = 4$
This implies that 64 is a cube of $4.$
Thus, -64 is the cube of $-4.$
View full question & answer→Question 93 Marks
Evaluate:
$\sqrt[3]{96}\times\sqrt[3]{144}$
Answer96 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 $24.$
View full question & answer→Question 103 Marks
Evaluate the following: $\Big\{(5^2+12^2)^{\frac{1}{2}}\Big\}^3$
AnswerTo evaluate the value of the given expression, we can proceed as follows: $\Big\{(5^2+12^2)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(25+144)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(169)^{\frac{1}{2}}\Big\}^3$
$=\Big\{\sqrt{(169)}\Big\}^3$
$=\Big\{\sqrt{(13\times13)}\Big\}^3$
$=\{13\}^3$
$13\times13\times13=2197$
View full question & answer→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 $3p + 2$ then $n^3$ also a number of the same type.
AnswerThree natural numbers of the form $(3p + 2)$ can be written by choosing
$p = 1, 2, 3...$ etc.
Let three such numbers be $5, 8$ and $11.$
Cubes of the three chosen numbers are:
$5^3= 125, 8^3= 512$ and $11^3= 1331$
Cubes of $5, 8$, and $11$ can be expressed as:
$125 = 3 \times 41 + 2,$ which is of the form $(3p + 2) for p = 41$
$512 = 3 \times 170 + 2,$ which is of the form $(3p + 2) for p = 170$
$1331 = 3 \times 443 + 2,$ which is of the form $(3p + 2) for p = 443$
Cubes of $5, 8$, and $11$ could be expressed as the natural numbers of the form $(3p + 2)$ for some natural number p. Hence, the statement is verified.
View full question & answer→Question 123 Marks
Making use of the cube root table, find the cube root $1100$
AnswerWe have: $1100 = 11 \times 100$
$\therefore\sqrt[3]{1100}$
$=\sqrt[3]{11\times100}$
$=\sqrt[3]{11}\times\sqrt[3]{100}$ 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 $10.323.$
View full question & answer→Question 133 Marks
Find The cube roots of the numbers $3048625, 20346417, 210644875, 57066625$ using the fact that. $3048625 = 3375 \times 729$
AnswerTo 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 $135.$
View full question & answer→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 $33.$
View full question & answer→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 $3025.$
View full question & answer→Question 163 Marks
Three numbers are in the ratio $1 : 2 : 3.$ The sum of their cubes is $98784$. Find the numbers.
AnswerLet the numbers are $= x, 2x$ and $3x$
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 $
$\text{x}^3=\frac{98784}{36}=2744$
$\text{x}=3_\sqrt{2744}$
$=3_\sqrt{2\times2\times2\times7\times7\times7}$
$= 2 \times 7 = 14 $
So, the numbers are ,
$x = 14 $
$2x = 2 \times 14 = 28 $
$3x = 3 \times 14 = 42$
View full question & answer→Question 173 Marks
Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer. −$5832$
AnswerIn 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 $-18.$
View full question & answer→Question 183 Marks
Find The cube roots of the numbers $3048625, 20346417, 210644875, 57066625$ using the fact that. $210644875 = 42875 \times 4913$
AnswerTo 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 $595.$
View full question & answer→Question 193 Marks
Evaluate: $\sqrt[3]{36}\times\sqrt[3]{384}$
Answer36 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 $24.$
View full question & answer→Question 203 Marks
Find The cube roots of the numbers $3048625, 20346417, 210644875, 57066625$ using the fact that. $57066625 = 166375 \times 343$
AnswerTo 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 $385.$
View full question & answer→Question 213 Marks
Find the cube root of each of the following natural numbers:$35937$
AnswerBy prime factorization method, $=3_\sqrt{35937}$ $=3_\sqrt{3\times3\times3\times11\times11\times11}$ $=3_\sqrt{3^3\times11^3}$ $=3\times11$ $=33$
View full question & answer→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 $3.$
AnswerThree 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.
View full question & answer→Question 233 Marks
Evaluate the following: $\sqrt[3]{\text{a}24}+\sqrt[3]{0.008}+\sqrt[3]{0.0064}$
AnswerTo 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 $3.6.$
View full question & answer→Question 243 Marks
Find the cube root of each of the following natural numbers: $48228544$
AnswerBy prime factorization method, $=3_\sqrt{48228544}$
$=3_\sqrt{2\times2\times2\times2\times2\times2\times7\times7\times7\times13\times13\times13}$
$=3_\sqrt{2^3\times2^3\times7^3\times13^3} = 2 \times 2 \times 7 \times 13 = 364$
View full question & answer→Question 253 Marks
Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer. $−2744000$
AnswerIn 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 $140.$
Thus, $-2744000$ is the cube of $-140.$
View full question & answer→Question 263 Marks
Which of the following numbers are cubes of negative integers. $−2197$
AnswerIn 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 $-13.$
View full question & answer→Question 273 Marks
Evaluate: $125\sqrt[3]{\text{a}^6}-\sqrt[3]{125\text{a}^6}$
AnswerProperty:
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:
$125\sqrt[3]{\text{a}^6}-\sqrt[3]{125\text{a}^6}$
$=125\sqrt[3]{\text{a}^6}\Big(\sqrt[3]{125}\times\sqrt[3]{\text{a}^6}\Big)$
$= 125 \times a^2- (5 \times a^2)$
$\Big(\because\sqrt[3]{\text{a}\times\text{a}\times\text{a}\}\times\text{\{a}\times\text{a}\times\text{a}\}}= \text{a} \times \text{a} = \text{a}^2 \text{and}\sqrt[3]{125}= \sqrt[]{5\times5\times5}=5\Big)$
$= 125a^2- 5a^2$
$= 120a^2$
View full question & answer→Question 283 Marks
Find the cube root of each of the following natural numbers:$1728$
AnswerBy prime factorization method, $=3_\sqrt{1728}$
$=3_\sqrt{2\times2\times2\times2\times2\times2\times3\times3\times3}$
$=3_\sqrt{2^3\times2^3\times3^3}$
$=2\times2\times3$
$=12$
View full question & answer→Question 293 Marks
Making use of the cube root table, find the cube root $5112$
AnswerBy 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:
$\sqrt[3]{9}=2.080$ and $\sqrt[3]{71}=4.141$
$\therefore\sqrt[3]{5112}$
$=2\times\sqrt[3]{9}\times\sqrt[3]{71}$
$=2\times2.080\times4.141$
$=17.227$ (upto three decimal places)
Thus, the required cube root is $17.227.$
View full question & answer→Question 303 Marks
Find the cube root of the following numbers: $-27 \times 2744$
AnswerProperty: 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 $\sqrt[3]{-27\times2744}$
$=-\sqrt[3]{27}\times\sqrt[3]{2744}$
$=-3\times6=-42$
View full question & answer→Question 313 Marks
Find the cube root of each of the following natural numbers: $33698267$
AnswerBy prime factorization method, $=3_\sqrt{33698267}$
$=3_\sqrt{17\times17\times17\times19\times19\times19}$
$=3_\sqrt{17^3 \times19^3} = 17 \times 19 = 323$
View full question & answer→Question 323 Marks
Which of the following numbers are cubes of negative integers. $−2744$
AnswerIn 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 $-14.$
View full question & answer→Question 333 Marks
Making use of the cube root table, find the cube root $7800$
AnswerWe have: $7800 = 78 \times 100 \therefore\sqrt[3]{7800}$
$=\sqrt[3]{78\times100}$
$=\sqrt[3]{78}\times\sqrt[3]{100}$ 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 $19.835.$
View full question & answer→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 $30.$
View full question & answer→Question 353 Marks
Find the cube root of each of the following natural numbers: $1157625$
AnswerBy prime factorization method, $=3_\sqrt{1157625}$
$=3_\sqrt{3\times3\times3\times5\times5\times5\times7\times7\times7}$
$=3_\sqrt{3^3\times5^3\times7^3} = 2 \times 5 \times 7 = 105$
View full question & answer→Question 363 Marks
Find the cube root of each of the following natural numbers: $17576$
AnswerBy prime factorization method, $=3_\sqrt{17576}$
$=3_\sqrt{2\times2\times2\times13\times13\times13}$
$=3_\sqrt{2^3\times13^3}$
$=2\times13$
$=26$
View full question & answer→Question 373 Marks
Which of the following are cubes of even natural numbers? $216, 512, 729, 1000, 3375, 13824$
AnswerWe 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 $13824.$
View full question & answer→Question 383 Marks
Evaluate the following: $\sqrt[3]{1000}+\sqrt[3]{0.008}-\sqrt[3]{0.125}$
AnswerTo 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 $9.7.$
View full question & answer→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.
AnswerThus, $243$ should be multiplied by $3$ to make it a perfect cube.On factorising $243$ into prime factors, we get:
$243 = 3 \times 3 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors, we get:
$243 = {3 \times 3 \times 3} \times 3 \times 3$
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 $(3 \times 3 = 9)$, the factors can be grouped into triples of equal factors such that no factor is left over.
View full question & answer→Question 403 Marks
Which of the following numbers are cubes of negative integers. $−1056$
AnswerIn 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.
View full question & answer→Question 413 Marks
Evaluate the following: $\Big\{(6^2+8^2)^{\frac{1}{2}}\Big\}^3$
AnswerTo evaluate the value of the given expression, we can proceed as follows: $\Big\{(6^2+8^2)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(36+64)^{\frac{1}{2}}\Big\}^3$
$=\Big\{(100)^{\frac{1}{2}}\Big\}^3$
$=\Big\{\sqrt{(100)}\Big\}^3$
$=\Big\{\sqrt{(10\times10)}\Big\}^3$
$=\{10\}^3$
$10\times10\times10=1000$
View full question & answer→Question 423 Marks
Find the cube roots of the following integers: $-2744000$
AnswerWe have, Cube root of -125 $=\sqrt[3]{-2744000}$
$=-\sqrt[3]{2744000}$ To find the cube root of $2744000$, we use the method of factorisation.
$2 \times 2 \times 5 \times 7 = 140$ 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.
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 $140$.
Hence, $=\sqrt[3]{-2744000}$
$= -\sqrt[3]{2744000}$
$=-140$
View full question & answer→Question 433 Marks
Making use of the cube root table, find the cube root $7000$
AnswerWe have: $700 = 70 \times 100$
$\therefore\sqrt[3]{7000}$
$=\sqrt[3]{7\times1000}$
$=\sqrt[3]{7}\times\sqrt[3]{1000}$ By the cube root table,
we have: $=\sqrt[3]{7}=1.913$ and $\sqrt[3]{1000}=10$
$\therefore\sqrt[3]{7000}$
$=\sqrt[3]{7}\times\sqrt[3]{1000}$
$=1.913\times10$
$=19.13$
View full question & answer→Question 443 Marks
Find the cube root of each of the following natural numbers: $2744$
AnswerBy prime factorization method, $=3_\sqrt{2774}$
$=3_\sqrt{2\times2\times2\times7\times7\times7}$
$=3_\sqrt{2^3\times7^3}$
$=2\times7$
$=14$
View full question & answer→Question 453 Marks
Making use of the cube root table, find the cube root $9800$
AnswerWe have:
$9800 = 98 \times 100$
$\therefore \sqrt[3]{9800}$
$=\sqrt[3]{98\times100}$
$=\sqrt[3]{98}\times\sqrt[3]{100}$
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 $21.40.$
View full question & answer→Question 463 Marks
Find the cube root of the following rational numbers: $\frac{686}{-3456}$
AnswerLet 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.) $=-\sqrt[3]{\frac{7^3}{2^6\times3^3}}$
$=\frac{-\sqrt[3]{7^3}}{\sqrt[3]{2^6\times^3}}$
$=\frac{{-7}}{\sqrt[3]{2^3\times2^3\times3^3}}$
$=\frac{{-7}}{\sqrt[3]{2\times2\times3}}$
$=\frac{-7}{12}$
$\Big(\because\sqrt[3]{\frac{\text{a}}{\text{b}}}=\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}\Big)$
View full question & answer→Question 473 Marks
Find the cube root of each of the following natural numbers: $134217728$
AnswerBy prime factorization method, $=3_\sqrt{134217728}$
$=3_\sqrt{2^{27}}$
$=2^9$
$=512$
View full question & answer→Question 483 Marks
Find the cube roots of the following integers:
$ -733571$
AnswerWe 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 $753 (9^3 < 753 < 10^3).$
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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