The value of a so that the sum of the cubes of the roots of the equat…

Let $\alpha$ and $\beta$ be the roots of the given equation. Then,

$\alpha+\beta=a$ and $\alpha \beta=2 a-3$

Let $S=\alpha^3+\beta^3$. Then,

$S =(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$

$\Rightarrow S =a^3-3 a(2 a-3)$

$\Rightarrow S =a^3-6 a^2+9 a$

$\Rightarrow \frac{d S}{d a}=3 a^2-12 a+9=3\left(a^2-4 a+3\right)=3(a-1)(a-3)$

Clearly, $\frac{d S}{d a}=0$ when a = 1, 3.

The changes in the signs of $\frac{d S}{d a}$ are shown in Figure.

Clearly, S is minimum at a = 3.