Let f(x) = x^3 and g(x) = 3^x,then the quadra | vedclass

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(D) Given $f(x) = x^3$ and $g(x) = 3^x$. We need to solve $(f \circ g)(x) = (g \circ f)(x)$ for $x 
eq 0$. $(f \circ g)(x) = f(g(x)) = f(3^x) = (3^x)

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$x^2 - 3 = 0$

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(D) Given $f(x) = x^3$ and $g(x) = 3^x$. We need to solve $(f \circ g)(x) = (g \circ f)(x)$ for $x 
eq 0$. $(f \circ g)(x) = f(g(x)) = f(3^x) = (3^x)^3 = 3^{3x}$. $(g \circ f)(x) = g(f(x)) = g(x^3) = 3^{x^3}$. Equating the two expressions: $3^{3x} = 3^{x^3}$. Since the bases are equal,we equate the exponents: $3x = x^3$. Rearranging the terms: $x^3 - 3x = 0$. Factoring out $x$: $x(x^2 - 3) = 0$. Since $x 
eq 0$,we have $x^2 - 3 = 0$. Thus,the required quadratic equation is $x^2 - 3 = 0$.

Let $f(x) = x^3$ and $g(x) = 3^x$,then the quadratic equation whose roots are solutions of the equation $(f \circ g)(x) = (g \circ f)(x)$ (for $x \neq 0$) is

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