[Algebra]
Problem: $\quad$ Let $a$ and $b$ be real numbers such that

$$
a^{3}-15 a^{2}+20 a-50=0 \quad \text { and } \quad 8 b^{3}-60 b^{2}-290 b+2575=0
$$

Compute $a+b$.

Solution: Each cubic expression can be depressed - that is, the quadratic term can be eliminated-by substituting as follows. Because $(a-p)^{3}=a^{3}-3 a^{2} p+3 a p^{2}-p^{3}$, setting $p=-\frac{(-15)}{3}=5$ and substituting $c+p=a$ transforms the expression $a^{3}-15 a^{2}+20 a-50$ into the equivalent expression $(c+5)^{3}-15(c+5)^{2}+20(c+5)-50$, which simplifies to $c^{3}-55 c-200$. Similarly, the substitution $d=b-\frac{5}{2}$ yields the equation $d^{3}-55 d=-200$. [This procedure, which is analogous to completing the square, is an essential step in the algebraic solution to the general cubic equation.]

Consider the function $f(x)=x^{3}-55 x$. It has three zeros, namely, 0 and $\pm \sqrt{55}$. Therefore, it has a relative maximum and a relative minimum in the interval $[-\sqrt{55}, \sqrt{55}]$. Note that for $0 \leq x \leq 5.5,|f(x)|<\left|x^{3}\right|<5.5^{3}=166.375$, and for $5.5<x \leq \sqrt{55}<8$, we have

$$
|f(x)|=\left|x^{3}-55 x\right|<x\left|x^{2}-55\right|<8\left(55-5.5^{2}\right)=198
$$

Because $f(x)$ is an odd function of $x$ (its graph is symmetric about the origin), we conclude that for $-\sqrt{55} \leq x \leq \sqrt{55},|f(x)|<198$. Therefore, for constant $m$ with $|m|>198$, there is a unique real number $x_{0}$ such that $f\left(x_{0}\right)=m$.

In particular, since $200>198$, the values of $c$ and $d$ are uniquely determined. Because $f(x)$ is odd, we conclude that $c=-d$, or $a+b=\frac{\mathbf{1 5}}{\mathbf{2}}$.
Set $a=x-b$ and substitute into the first equation. We get

$$
\begin{aligned}
(x-b)^{3}-15(x-b)^{2}+20(x-b)-50 & =0 \\
-b^{3}+b^{2}(3 x-15)+b\left(-3 x^{2}+30 x-20\right)+\left(x^{3}-15 x^{2}+20 x-50\right) & =0 \\
8 b^{3}+b^{2}(-24 x+120)+b\left(24 x^{2}-240 x+160\right)-8\left(x^{3}-15 x^{2}+20 x-50\right) & =0 .
\end{aligned}
$$

If we equate coefficients, we see that

$$
\begin{aligned}
-24 x+120 & =-60 \\
24 x^{2}-240 x+160 & =-290 \\
-8\left(x^{3}-15 x^{2}+20 x-50\right) & =2575
\end{aligned}
$$

are all satisfied by $x=\frac{15}{2}$. This means that any real solution $b$ to the second equation yields a real solution of $\frac{15}{2}-b$ to the first equation. We can follow the reasoning of the previous solution to establish the existence of exactly one real solution to the second cubic equation. Thus $a$ and $b$ are unique, and their sum is $\left(\frac{15}{2}-b\right)+b=\frac{\mathbf{1 5}}{\mathbf{2}}$.

Final answer: $\frac{15}{2}$