Question

If $\alpha$ and $\beta$ be two zeros of the quadratic polynomial $a{x}^2+bx+c$, then evaluate:${\alpha }^3+{\beta }^3$

• A. $\frac{abc-b^3}{a^2}$
• B. $\frac{3abc-b^3}{a^3}$
• C. $\frac{abc-b^3}{2a^2}$
• D. $\frac{3abc-b^3}{2a^3}$

Answer 1

Answer: (B)

$\frac{3abc-b^3}{a^3}$

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 $=-\frac{b}{a}$

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

Poduct of the zeros$=\frac{c}{a}$

$\alpha \beta =>\frac{c}{a}$

Now,

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

Squaring both sides,

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

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

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

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

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

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

Now,

${\alpha }^3+{\beta }^3$

Using,$a^3+b^3=(a+b)(a^2+b^2-ab)$

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

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

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

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

$=\frac{3abc-b^3}{a^3}$