It's The Real Deal. Topic: Algebra. Level: Olympiad.

Problem: (360 Problems for Mathematical Contests - 1.1.18) Find the real solutions to the system of equations

$ \frac{1}{x}+\frac{1}{y} = 9 $
$ \left(\frac{1}{\sqrt[3]{x}}+\frac{1}{\sqrt[3]{y}}\right) \left(1+\frac{1}{\sqrt[3]{x}}\right) \left(1+\frac{1}{\sqrt[3]{y}}\right) = 18 $.

Solution: To simplify things, make the substitution

$ a = \frac{1}{\sqrt[3]{x}} $, $ b = \frac{1}{\sqrt[3]{y}} $.

We now have the system

$ a^3+b^3 = (a+b)(a^2-ab+b^2) = 9 $
$ (a+b)(1+a)(1+b) = (a+b)(1+a+b+ab) = 18 $.

Letting $ p = a+b $, $ q = ab $, it becomes

$ p(p^2-3q) = 9 $ (1)
$ p(1+p+q) = 18 $ (2)

Adding three times (2) to (1), we have

$ p(p^2+3p+3) = 63 $,

or

$ (p-3)(p^2+6p+21) = 0 $ (3)

Since the discriminant of $ p^2+6p+21 $ is $ 6^2-4(21) < 0 $, the only real solution is $ p = 3 $. Then from (1), we have

$ 3(9-3q) = 9 \Rightarrow q = 2 $.

So $ ab = 2 \Rightarrow a = \frac{2}{b} $. So

$ p = \frac{2}{b}+b = 3 \Rightarrow b^2-3b+2 = (b-1)(b-2) = 0 $.

Then $ b = 1, 2 $ giving the solutions

$ (a,b) = (1,2); (2,1) $, which become

$ (x,y) = \left(1, \frac{1}{8}\right); \left(\frac{1}{8},1\right) $. QED.

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Comment: The only really hard part about this problem was the ugliness of the original problem. Making the right substitutions reduces it to simple algebra and solving a cubic (with an easy root if you saw it).

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Practice Problem: Prove that the equation $ a^2+4b^2+4ab+4a+8b+5 = 0 $ has no solution in the reals.