Question

If the equation $x^3+2x^2-4x+5=0$ 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)

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