22. Suppose that $x_{1}, x_{2}$ and $x_{3}$ are the three roots of $(11-x)^{3}+(13-x)^{3}=(24-2 x)^{3}$. Find the value of $x_{1}+x_{2}+x_{3}$.
22. Answer: 36
Let $a=11-x$ and $b=13-x$. Hence the equation becomes
$$
a^{3}+b^{3}=(a+b)^{3}
$$

Using the binomial expansion
$$
(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b),
$$

Equation (1) becomes
$$
a b(a+b)=0
$$

Thus the solution of (1) becomes
$$
a=0, b=0 \text { or } a+b=0
$$
or equivalently, the solution of the original equation becomes $x=11,12$ or 13 .
Hence $x_{1}+x_{2}+x_{3}=11+12+13=36$.