Exercise 3. Let $a$ be a real number such that $(a-1)(a-2)(a-3)+a^{2}(6-a)=0$. Calculate $11a$.
Solution of the exercise $3(a-1)(a-2)=a(a-2)-(a-2)=a^{2}-2 a-a+2=a^{2}-3 a+2$ therefore $(a-1)(a-2)(a-3)=a\left(a^{2}-3 a+2\right)-3\left(a^{2}-3 a+2\right)=a^{3}-3 a^{2}+2 a-3 a^{2}+9 a-6=$ $a^{3}-6 a^{2}+11 a-6$. Moreover, $a^{2}(6-a)=6 a^{2}-a^{3}$, so $11 a-6=0$, and finally $11 a=6$.