Cube of Polynomials

Let $a = 2 \times 3^{x}$ and $b = 9^{x} - 3$ . Then the given equation can be written as

$a^{3} + b^{3} = (a + b)^{3} \Longrightarrow a^{3} + b^{3} = a^{3} + b^{3} + 3ab(a + b) \Longrightarrow 3ab(a + b) = 0$ ,

which is satisfied whenever $a = 0, b = 0$ or $a + b = 0$ . Now $a = 2 \times 3^{x} \ne 0$ for all $x$ , but $b = 9^{x} - 3 = 0$ for $x = \dfrac{1}{2}$ .

Finally, $a + b = 2 \times 3^{x} + 9^{x} - 3 = 0 \Longrightarrow y^{2} + 2y - 3 = (y + 3)(y - 1) = 0$ where $y = 3^{x}$ . But as $3^{x} \gt 0$ for all $x$ we can only have $y = 3^{x} = 1 \Longrightarrow x = 0$ .

Thus the two solutions are $\boxed{0, \dfrac{1}{2}}$ .