Question

Calculate $\frac{1}{x_1^3}+\frac{1}{x_2^3},$, where $x_1$ and $x_2$ are roots of the equation $2x^2-3ax-2=0$

Answer 1

$2x^2-3ax-2=0$

$\text{Sum}=x_1+x_2=\frac{3a}{2}$

$\text{Product}=x_1\times x_2=-1$

$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_1^3+x_2^3}{(x_1{x}_2{)}^3}$

$=\frac{(x_1+x_2{)}^3-3x_1{x}_2(x_1+x_2)}{(x_1{x}_2{)}^3}$

$=\frac{(\frac{3a}{2}{)}^3-3(-1\left)\right(\frac{3a}{2})}{-1^2}$

$=\frac{\frac{27a^3}{8}+\frac{9a}{2}}{-1}$

$=\frac{-(27a^3+36a)}{8}$