Question

If $\alpha$ and $\beta$ be two zeros of the quadratic polynomial $a{x}^2+bx+c$, then $\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}$ is equal to

• A. $\frac{3abc-b^3}{c^3}$
• B. $\frac{abc-b^3}{c^3}$
• C. $\frac{ac-b^3}{c^3}$
• D. $\frac{3abc-a^3}{c^3}$

Answer 1

The correct option is A $\frac{3abc-b^3}{c^3}$

Given quadratic polynomial is $f\left(x\right)=a{x}^2+bx+c$
The zeroes of the polynomial are $\alpha \beta$
Sum of the zeros $=-\frac{b}{a}$
$\alpha +\beta =>-\frac{b}{a}$
Product of the zeros$=\frac{c}{a}$
$\alpha \beta =>\frac{c}{a}$
Now,
$\alpha +\beta =>-\frac{b}{a}$
Squaring both sides,
$=>(\alpha +\beta {)}^2=-(\frac{b}{a}{)}^2$

$=>{\alpha }^2+{\beta }^2+2\alpha \beta =\frac{b^2}{a^2}$

$=>{\alpha }^2+{\beta }^2+2\left(\frac{c}{a}\right)=\frac{b^2}{a^2}$

$=>{\alpha }^2+{\beta }^2=\frac{b^2}{a^2}-\frac{2c}{a}$
Now,
$\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}$ $=\frac{{\beta }^3+{\alpha }^3}{{\alpha }^3{\beta }^3}$

$=\frac{(\alpha +\beta )({\alpha }^2+{\beta }^2-\alpha \beta )}{(\alpha \beta {)}^3}$

$=\frac{\left(\frac{-b}{a}\right)(\frac{b^2}{a^2}-\frac{2c}{a}-\frac{c}{a})}{(\frac{c}{a}{)}^3}$

$=\frac{\left(\frac{-b}{a}\right)(\frac{b^2}{a^2}-\frac{3c}{a})}{(\frac{c}{a}{)}^3}$

$=\frac{\left(\frac{-b}{a}\right)\left(\frac{b^2-3ac}{a^2}\right)}{(\frac{c}{a}{)}^3}$

$=\frac{(-b)\left(\frac{b^2-3ac}{a^3}\right)}{(\frac{c}{a}{)}^3}$

$=\frac{\left(\frac{-b^3+3bac}{a^3}\right)}{\frac{c^3}{a^3}}$

$=\frac{-b^3{a}^3+3b{a}^{4}c}{c^3{a}^3}$

$=\frac{3bac-b^3}{c^3}$