1 Marks Question

Question 11 Mark

Write the value of $25^3 - 75^3+ 50^3.$

Answer

The given expression is $25^3-75^3+50^3$ Let $a =25, b=-75$ and $c =50$.
Then the given expression becomes $25^3-75^3+50^3= a ^3+ b ^3+ c ^3$
Note that $a+b+c=25+(-75)+50=25-75+50=0$
Recall the formula $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
when $a+b+c=0$, this becomes $a^3+b^3+c^3-3 a b c$
$=0 \cdot\left(a^2+b^2+c^2-a b-b c-c a\right)=0 a^3+b^3+c^3=3 a b c$ So,
we have the new formula $a^3+b^3+c^3=3 a b c$,
when $a+b+c=0$ Using the above formula,
the value of the given expression is $a ^3+ b ^3+ c ^3=3 abc 25^3-75^3+50^3$
$=3 \cdot(25) \cdot(-75) \cdot(50) 25^3-75^3+50^3=-281250$

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Question 21 Mark

Write the value of $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3.$

Answer

The given expression is $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3$
Let $\text{a}=\frac{1}{2},\text{b}=\frac{1}{3}$ and $\text{c}=-\frac{5}{6}.$
Then the given expression becomes $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=\text{a}^3+\text{b}^3+\text{c}^3$
Note that $\text{a}+\text{b}+\text{c}=\frac{1}{2}+\frac{1}{3}+\Big(\frac{5}{6}\Big)$
$=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$
$=0$ Recall the formula $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
when $a + b + c = 0$, this becomes
$a^3 + b^3 + c^3 - 3abc = 0.(a^2 + b^2 + c^2 - ab - bc - ca) = 0 a^3 + b^3 + c^3 = 3abc$
So, we have the new formula $a^3 + b^3 + c^3 = 3abc$, when $a + b + c = 0$
Using the above formula, the value of the given expression is $a^3 + b^3 + c^3 = 3abc$
$\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=3.\Big(\frac{1}{2}\Big).\Big(\frac{1}{3}\Big).\Big(-\frac{5}{6}\Big)$
$\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=\Big(-\frac{5}{12}\Big)$

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Question 31 Mark

Write the value of $30^3 + 20^3 - 50^3.$

Answer

The given expression is $30^3+20^3-50^3$ Let $a=30, b=20$ and $c=-50$.
Then the given expression becomes $30^3+20^3-50^3=a^3+b^3+c^3$
Note that $a+b+c=30+20+(-50)=30+20-50=0$
Recall the formula $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
when $a+b+c=0$, this becomes $a^3+b^3+c^3-3 a b c=0 \cdot\left(a^2+b^2+c^2-a b-b c-c a\right)=0 a^3+b^3+c^3=3 a b c$
So, we have the new formula $a^3+b^3+c^3=3 a b c$, when $a+b+c=0$
Using the above formula, the value of the given expression is $a ^3+ b ^3+ c ^3=3 abc 30^3+20^3-50^3$
$=3 \cdot(30) \cdot(20) \cdot(-50) 30^3+20^3-50^3=-90000$

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Question 41 Mark

Write the value of $48^3-30^3-18^3$.

Answer

The given expression is
$48^3-30^3-18^3$
Let $\mathrm{a}=48, \mathrm{~b}=-30$ and $\mathrm{c}=-18$. Then the given expression becomes $48^3-30^3-18^3=\mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3$
Note that
$a+b+c=48+(-30)+(-18)$
$=48-30-18$
$=0$
Recall the formula
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
when $a+b+c=0$, this becomes
$a^3+b^3+c^3-3 a b c=0 \cdot\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=0$
$a^3+b^3+c^3=3 a b c$
So, we have the new formula
$a^3+b^3+c^3=3 a b c \text {, when } a+b+c=0$
Using the above formula, the value of the given expression is
$a^3+b^3+c^3=3 a b c$
$48^3-30^3-18^3=3 \cdot(48) \cdot(-30) \cdot(-18)$
$48^3-30^3-18^3=77760$

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Question 51 Mark

Factorize: $x^2-1-2 a-a^2$.

Answer

$(x+a+1)(x-a-1)$

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Question 61 Mark

Factorize: $x^4+x^2+25$.

Answer

$\left(x^2+3 x+5\right)\left(x^2-3 x+5\right)$

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Question 71 Mark