Task A-2.5. (8 points)

Let $x_{1}, x_{2}$ be the distinct solutions of the equation $2 x^{2}-3 x+4=0$. Calculate $\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}}$.
## Solution.

From Vieta's formulas, we get $x_{1}+x_{2}=\frac{3}{2}, x_{1} \cdot x_{2}=2$.

It holds that $\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}}=\frac{x_{1}^{3}+x_{2}^{3}}{x_{1}^{3} \cdot x_{2}^{3}}=\frac{\left(x_{1}+x_{2}\right)^{3}-3 \cdot x_{1} x_{2} \cdot\left(x_{1}+x_{2}\right)}{\left(x_{1} x_{2}\right)^{3}}=\frac{\left(\frac{3}{2}\right)^{3}-3 \cdot 2 \cdot \frac{3}{2}}{2^{3}}=-\frac{45}{64}$.

For 7 points: good method with a calculation error.

For 8 points: correct result and method.