Question

If the equation $x^{3}+2 x^{2}-4 x+5=0$ has roots $\alpha, \beta$ and $\gamma$, then the value of $\frac{\left(\alpha^{3}+5\right)\left(\beta^{3}+5\right)\left(\gamma^{3}+5\right)}{13 \alpha \beta \gamma}$ is equal to:

• A. 5
• B. 8
• C. 12
• D. 15

Answer 1

Answer: (B)

$x^{3}+2 x^{2}-4 x+5 \equiv(x-\alpha)(x-\beta)(x-\gamma)$
put $x=2$, we get
$13=(2-\alpha)(2-\beta)(2-\gamma)...$(1)
Also, $\frac{\left(\alpha^{3}+5\right)\left(\beta^{3}+5\right)\left(\gamma^{3}+5\right)}{13 \alpha \beta \gamma}$ $=\frac{2 \alpha \cdot(2-\alpha) \cdot 2 \beta \cdot(2-\beta) \cdot 2 \gamma \cdot(2-\gamma)}{13 \alpha \beta \gamma}$
$=\frac{8}{13}(13)=8$