It is given that roots of the equation $x^3-6 x^2+11 x-6=0$ are $\alpha, \beta, \gamma$.
Now, to find the equation whose roots are $\alpha^2, \beta^2, \gamma^2$, put $\alpha^2=x \Rightarrow \alpha=\sqrt{x}$.
Since, $\alpha$ is the root of the given equation, so
$$
\begin{aligned}
& x^{3 / 2}-6 x+11 x^{1 / 2}-6=0 . \\
& \Rightarrow \quad x^{1 / 2}(x+11)=6(x+1)
\end{aligned}
$$

On squaring both sides, we get
$$
\begin{aligned}
& & x(x+11)^2=36(x+1)^2 \\
\Rightarrow & & x\left[x^2+22 x+121\right]=36\left[x^2+2 x+1\right] \\
\Rightarrow & & x^3+22 x^2+121 x=36 x^2+72 x+36 \\
\Rightarrow & x^3-14 x^2+49 x-36 & =0
\end{aligned}
$$