Find all real numbers $a, b$ such that the polynomial $(X-1)^{2}$ divides the polynomial $a X^{4}+$ $b X^{3}+1$.
Let $\mathrm{P}(\mathrm{x})=\mathrm{a} x^{4}+\mathrm{b} x^{3}+1$.

First SOLUTION. We have seen that $(x-1)^{2}$ divides $P(x)$ if and only if $P(1)=0$ and $P^{\prime}(1)=0$. Now, $P(1)=a+b+1$ and $P^{\prime}(1)=4 a+3 b$. Solving $a+b+1=0$ and $4 a+3 b=0$ gives $b=-4$ and $a=3$.

SECOND SOLUTION. We perform the Euclidean division of $\mathrm{a} x^{4}+\mathrm{b} x^{3}+1$ by $(x-1)^{2}$. After calculations, we find

$$
a x^{4}+b x^{3}+1=\left(x^{2}-2 x+1\right)\left(a x^{2}+(b+2 a) x+2 b+3 a\right)+(3 b+4 a) x+1-2 b-3 a
$$

Thus $(x-1)^{2}$ divides $a x^{4}+b x^{3}+1$ if and only if $3 b+4 a=0$ and $2 b+3 a=1$, which is to say if and only if $b=-4$ and $a=3$.

THIRD SOLUTION. Suppose that $a x^{4}+b x^{3}+1=(x-1)^{2} P(x)$. The polynomial $P(x)$ is monic of degree 2, and we can thus write $P(x)=x^{2}+u x+v$. By expanding the equality $a x^{4}+b x^{3}+1=(x-1)^{2}\left(x^{2}+u x+v\right)$, identifying the coefficients, and solving the system of two equations, we find $b=-4$ and $a=3$. Conversely, we verify that these values are suitable.