Question

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

$\therefore (\alpha +\beta )(\beta +\gamma )(\gamma +\alpha )\ge 8(\alpha \beta \gamma )$
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
$\therefore 8-4m+24-4n=0\Rightarrow 32=4(m+n)\Rightarrow m+n=8$