Question

If $\alpha and\beta$ are the roots of the equation ax²+bx+c=0, then the value of ${\alpha }^3+{\beta }^3$ is

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

Answer 1

Answer: (C)

Given : $\alpha$ & $\beta$ are roots of equation
ax² + bx + c = 0
$\therefore \alpha +\beta =-\frac{b}{a}\&\alpha \beta =\frac{c}{a}$
Now, ${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$
$\Rightarrow {\alpha }^3+{\beta }^3=(-\frac{b}{a}{)}^3-3\frac{c}{a}(-\frac{b}{a})$
$\Rightarrow {\alpha }^3+{\beta }^3=-\frac{b^2}{a^3}+\frac{3bc}{a^2}$
$\Rightarrow {\alpha }^3+{\beta }^3=\frac{-b^3+3abc}{a^3}$