20. Let $x$ be a non-zero real number such that $\sqrt[5]{x^{3}+20 x}=\sqrt[3]{x^{5}-20 x}$. Find the product of all possible values of $x$.
(2 marks)
Let $x$ be a non-zero real number such that $\sqrt[5]{x^{3}+20 x}=\sqrt[3]{x^{5}-20 x}$. Find the product of all possible values of $x$.
(2 marks)
20. -5
20. Let \( z = \sqrt[5]{x^3 + 20x} = \sqrt[3]{x^5 - 20x} \). Then we have \( z^5 = x^3 + 20x \) and \( z^3 = x^5 - 20x \). Adding, we have \( z^5 + z^3 = x^5 + x^3 \), or \( z^5 - x^5 + z^3 - x^3 = 0 \). Hence we have
\[
0 = (z - x)\left(z^4 + z^3 x + z^2 x^2 + z x^3 + x^4 + z^2 + z x + x^2\right).
\]

As \( x \) is non-zero, \( z \) is also non-zero. Thus we have \( z^2 + z x + x^2 = \left(z - \frac{x}{2}\right)^2 + \frac{3}{4} x^2 > 0 \) and \( z^4 + z^3 x + z^2 x^2 + z x^3 + x^4 = z^2 \left(z + \frac{x}{2}\right)^2 + x^2 \left(x + \frac{z}{2}\right)^2 + \frac{1}{2} z^2 x^2 > 0 \). Therefore the second factor on the right-hand side of the above equation is positive. This forces \( z = x \) and hence we get \( x = \sqrt[5]{x^3 + 20x} \), or \( 0 = x^5 - x^3 - 20x = x(x^2 + 4)(x^2 - 5) \). The possible non-zero real values of \( x \) are thus \( \pm \sqrt{5} \) and so the answer is -5.