Question

If $\alpha, \beta, \gamma$ are positive real roots of the equation $x^3-m^2+12 x-4 n=0(m, n \in R)$ such that $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=8 \alpha \beta \gamma$, then the value of $(m+n)$ is equal to

• A. 12
• B. 16
• C. 0
• D. 8

Answer 1

Answer: (D)

Given $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)=8 \alpha \beta \gamma$

$\therefore (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) \geq 8(\alpha \beta \gamma)$
Hence $\alpha=\beta=\gamma$
$\therefore \alpha \beta+\beta \gamma+\gamma \alpha=12 \Rightarrow 3 \alpha^2=12 \Rightarrow \alpha=2$
Put $ \alpha=2$ in the given equation
$\therefore 8-4 m+24-4 n=0 \Rightarrow 32=4(m+n) \Rightarrow m+n=8$