Question

Without solving the equation $3x^2-5x-2=0$. Find the sum of the cubes of its roots.

Answer 1

$\Rightarrow$ Let $\alpha$ and $\beta$ roots of the equation $3x^2-5x-2=0$

Here, $a=3,b=-5,c=-2$

$\Rightarrow$ $\alpha +\beta =\frac{-b}{a}=\frac{-(-5)}{3}=\frac{5}{3}$ ----- ( 1 )

$\Rightarrow$ $\alpha \beta =\frac{c}{a}=\frac{-2}{3}$ ----- ( 2 )

Now,

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

$\Rightarrow$ ${\left(\frac{5}{3}\right)}^3={\alpha }^3+{\beta }^3+3\alpha \beta (\alpha +\beta )$

$\Rightarrow$ $\frac{125}{27}={\alpha }^3+{\beta }^3+3\times \frac{(-2)}{3}\times \frac{5}{3}$

$\Rightarrow$ $\frac{125}{27}={\alpha }^3+{\beta }^3-\frac{10}{3}$

$\Rightarrow$ ${\alpha }^3+{\beta }^3=\frac{125}{27}+\frac{10}{3}$

$\therefore$ ${\alpha }^3+{\beta }^3=\frac{215}{27}$