Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Slovenia

Let $a$ and $b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b} + \frac{a-b}{a+b} = 6$ .
Find the value of the expression $\frac{a^3 + b^3}{a^3 - b^3} + \frac{a^3 - b^3}{a^3 + b^3}$ .

Solution

The equality
$6 = \frac{a+b}{a-b} + \frac{a-b}{a+b} = \frac{2a^2 + 2b^2}{a^2 - b^2}$
implies $6a^2 - 6b^2 = 2a^2 + 2b^2$ , or $4a^2 = 8b^2$ . From here we get $a = \pm b\sqrt{2}$ . Now, we can conclude that
$\frac{a^3 + b^3}{a^3 - b^3} + \frac{a^3 - b^3}{a^3 + b^3} = \frac{2a^6 + 2b^6}{a^6 - b^6} = \frac{(2 \cdot 2^3 + 2)b^6}{(2^3 - 1)b^6} = \frac{18}{7}.$

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