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

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

Since $x_1,x_2$ are the roots of the given equation

$x_1+x_2=-(-3a/2)=3a/2$

$x_1{x}_2=-2/2=-1$

$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_1^3+x_2^3}{x_1^2{x}_2^3}=\frac{x_1+x_2(x_1+x_2-x_1{x}_2)}{x_1^3{x}_2^3}$

Now, $x_1^2+x_2^2=(x_1+x_2{)}^2-2x_1+x_2=(\frac{9a^2}{4}+2)$

$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_1+x_2(x_1+x_2-x_1{x}_2)}{x_1^3{x}_2^3}=\frac{3a}{2}(\frac{9a^2}{4}+2)=\frac{3a(9a^2+8)}{8}$