We have given that $\alpha$ is non-real root of $x^7=1$
$$
\therefore \alpha^7=1 \text { and } \alpha \neq 1
$$
Now, $\alpha(1+\alpha)\left(1+\alpha^2+\alpha^4\right)$
$$
\begin{aligned}
& =\alpha\left(1+\alpha^2+\alpha^4+\alpha+\alpha^3+\alpha^5\right) \\
& =\alpha+\alpha^3+\alpha^5+\alpha^2+\alpha^4+\alpha^6 \\
& =\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6 \\
& =\frac{\alpha\left(1-\alpha^6\right)}{1-\alpha} \quad[\because \alpha \neq 1] \\
& =\frac{\alpha-\alpha^7}{1-\alpha}=\frac{\alpha-1}{1-\alpha} \quad\left[\because \alpha^7=1\right] \\
& \therefore \quad \alpha(1+\alpha)\left(1+\alpha^2+\alpha^4\right)=-1
\end{aligned}
$$