Answer:
Option C
Explanation:
Let \alpha,\beta ,\gamma are roots of equation
x^{3}-ax^{2}+ax-1=0 ..........(i)
\therefore \alpha+\beta+\gamma=a
\alpha\beta +\beta\gamma+\alpha\gamma=a
\alpha\beta \gamma=-1
Cubic equation whose roots \alpha^{2}, \beta^{2}, \gamma ^{2} is
x^{2}-(\alpha^{2}+\beta^{2}+ \gamma^{2})x^{2}+(\alpha^{2} \beta^{2}+\beta^{2}\gamma^{2}+\alpha^{2} \gamma^{2})x-\alpha ^{2}\beta^{2} \gamma ^{2}=0 ........(ii)
Equi .(i) and (ii) are identical.
\therefore\frac{a}{\alpha^{2}+\beta^{2}+ \gamma^{2}}=\frac{a}{\alpha^{2}\beta^{2}+\beta^{2}\gamma^{2}+ \alpha^{2}\gamma^{2}}=\frac{1}{\alpha^{2}\beta^{2} \gamma^{2}}
a= \alpha ^{2}+\beta ^{2}+\gamma ^{2} [ \alpha \beta \gamma=-1]
a= (\alpha +\beta +\gamma )^{2}-2(\alpha \beta + \beta \gamma +\gamma \alpha )
a= a^{2}-2a \Rightarrow a^{2}=3a
\Rightarrow a=3 [ \because a is non-zero real ]
