a) Determine the polynomial $\mathrm{Q}$ such that for any real $\alpha, \mathrm{Q}(\cos \alpha)=\cos 3 \alpha$.

b) Let $a, b, c$ be any real numbers, $P(x)=4 x^{3}+a x^{2}+b x+c$ and $M$ the maximum value of $|P(x)|$ on $[-1,1]$. Show that $M \geqslant 1$. In what case do we have equality?
$\cos 2 \alpha=\cos (\alpha+\alpha)=\cos ^{2} \alpha-\sin ^{2} \alpha=2 \cos ^{2} \alpha-1$, and $\cos 3 \alpha=\cos (2 \alpha+\alpha)=\left(2 \cos ^{2} \alpha-\right.$ 1) $\cos \alpha-(2 \sin \alpha \cos \alpha) \sin \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, so $Q(x)=4 x^{3}-3 x$.

This polynomial is useful for question b). Indeed, it equals 1 when $3 \alpha$ is a multiple of $2 \pi$, i.e., when $x=\cos \alpha=1$ or $\frac{-1}{2}$, and equals -1 when $3 \alpha$ is an odd multiple of $\pi$, i.e., when $x=\cos \alpha=\frac{1}{2}$ or -1. For any other value of $x=\cos \alpha \in[-1,1]$, $Q(x)$ is between -1 and 1. Therefore, for $a=0, b=-3, c=0, M=1$.

If a polynomial $P(x)$ satisfied for all $x \in[-1,1]|P(x)|0,(P-Q)\left(\frac{-1}{2}0\right.$ and $(P-Q)(1)\frac{1}{2}$ in which case $\mathrm{P}(-1)0$ : contradiction in both cases. The polynomial $Q$ is thus the unique case of equality.