Explanation

(c)
Let $P(n) = x\left( x^{n - 1} - n\alpha^{n - 1} \right) + \alpha^{n}(n - 1) = (x - \alpha)^{2}g(x)$
$P(1) \equiv 0$ is true.
Let $P(k)$ is true.
$ie,\ x\left( x^{k - 1} - k\alpha^{k - 1} \right) + \alpha^{k}(k - 1) = (x - \alpha)^{2}g(x)$
Now, $P(k + 1) \equiv x\left\lbrack x^{k} - (k + 1)\alpha^{k} \right\rbrack{+ \alpha}^{k + 1}(k)$
$\equiv (x - \alpha)^{2}\lbrack xg(x) + k\alpha^{k - 1}\rbrack$ (True)
So, holds for all $n \in N$.