20. [11] Compute $\sqrt[4]{5508^{3}+5625^{3}+5742^{3}}$, given that it is an integer.
Answer:
855
Solution: Let $a=5625=75^{2}$ and $b=117$. Then we have
$$
5508^{3}+5265^{3}+5742^{3}=(a-b)^{3}+a^{3}+(a+b)^{3}=3 a^{3}+6 a b^{2}=3 a\left(a^{2}+2 b^{2}\right) .
$$

We have $3 a=3^{3} \cdot 5^{4}$, so $a^{2}+2 b^{2}=3^{4} \cdot\left(625^{2}+2 \cdot 19^{2}\right)$ should be 3 times a fourth power. This means
$$
625^{2}+2 \cdot 19^{2}=3 x^{4}
$$
for some integer $x$. By parity, $x$ must be odd, and also $x^{2} \sqrt{3} \approx 625$. Approximating $\sqrt{3}$ even as 2 , we get $x$ should be around 19. Then $x=17$ is clearly too small, and $x=21$ is too big. (You can also check mod 7 for this latter one.) Thus, $x=19$. The final answer is then
$$
3^{2} \cdot 5 \cdot 19=855 \text {. }
$$