Question

If $\alpha \&\beta$ are roots if the equation $x^2+5x-5=0$, then evaluate $\frac{1}{{(\alpha +1)}^3}+\frac{1}{{(\beta +1)}^3}$

• A. $\frac{7}{27}$
• B. $\frac{5}{27}$
• C. $\frac{4}{27}$
• D. $\frac{1}{27}$

Answer 1

Answer: (C)

$\frac{4}{27}$

$x^2+5x-5=0$

Comparing with $a{x}^2+bx+c=0$

$a=1$ $b=5$ $C=-5$

the roots are $\alpha$ & $\beta$

$\alpha +\beta =\frac{-b}{a}=-5$ $\alpha \beta =\frac{c}{a}=-5$

here $\underset{a}{\underset{}{\frac{1}{(\alpha +1{)}^3}}}+\underset{b}{\underset{}{\frac{1}{(\beta +1{)}^3}}}$

let $\alpha +1=a$

$\beta +1=b$

so $\frac{1}{a^3}+\frac{1}{b^3}$

$=\frac{b^3+a^3}{(ab{)}^3}$

$=\frac{(a+b)(a^2+b^2-ab)}{(ab{)}^3}$

$=\frac{(a+b)\left[\right(a+b{)}^2-3ab]}{(ab{)}^3}$

$=\frac{(\alpha +1+\beta +1)\left[\right(\alpha +1+\beta +1{)}^2-3(\alpha +1)(\beta +1)]}{\left[\right(\alpha +1\left)\right(\beta +1){]}^3}$

$=\frac{(\alpha +\beta +2)\left[\right(\alpha +\beta +2{)}^2-3(\alpha +\beta +\alpha \beta +1)]}{(\alpha \beta +\alpha +\beta +1{)}^3}$

$=\frac{(-5+2)\left[\right(-5+2{)}^2-3(-5-5+1)]}{(-5-5+1{)}^3}$

$=\frac{(-3)\left[\right(-3{)}^2-3(-9)]}{(-9{)}^3}$

$=\frac{36\times (-3)}{(-9)\times (-9)\times (-9)}=\frac{4}{27}$