Let $\alpha, \beta, \gamma$ are roots of equation
$$
\begin{aligned}
x^3-a x^2+a x-1 & =0 \\
\alpha+\beta+\gamma & =a \\
\alpha \beta+\beta \gamma+\alpha \gamma & =a \\
\alpha \beta \gamma & =-1
\end{aligned}
$$
Cubic equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is
$$
\begin{array}{r}
x^2-\left(\alpha^2+\beta^2+\gamma^2\right) x^2+\left(\alpha^2 \beta^2+\beta^2 \gamma^2+\alpha^2 \gamma^2\right) x \\
-\alpha^2 \beta^2 \gamma^2=0 .
\end{array}
$$
Eqs. (i) and (ii) are identical.
$$
\therefore \frac{a}{\alpha^2+\beta^2+\gamma^2}=\frac{a}{\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2}=\frac{1}{\alpha^2 \beta^2 \gamma^2}
$$

$$
\begin{array}{rlr}
a & =\alpha^2+\beta^2+\gamma^2 & {[\alpha \beta \gamma=-1]} \\
a & =\left(\alpha+\beta+\gamma^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)\right. \\
a & =a^2-2 a \Rightarrow \quad a^2=3 a \\
\Rightarrow \quad a & =3 & {[\because a \text { is non-zero reaI }]}
\end{array}
$$