Let f ( x ) = x 3 and g ( x ) = 3 x , then the quadratic equation whose roots are solutions of the equation…

Let $f\left(x\right)=x^3$ and $g\left(x\right)=3^x$

$(f\circ g)\left(x\right)=(g\circ f)\left(x\right)$             (for $x\ne 0$) 

$(f\circ g )\left(x\right) = f( g (x) ) = f( 3x) = 3^{3x}$

$(g \circ f)\left(x\right) = g( f (x) ) = g( x^3) = 3^{x^3}$

Equating the powers with same base

$x^3 - 3x = 0$

$x(x^2 - 3) = 0$

$x^2 - 3 = 0$, $\left(x \ne 0 \right)$