I am trying to multiply $f(x)$ with the given equation to get $g(x)$ but I am not getting the right answer. Any help would be appreciated. The link to the question is posted.
Define functions $f(x)$ and $g(x)$ by $$ f(x) = \frac{\sqrt[3]{x} - 2}{x-8}, \quad g(x) = \frac1{\sqrt[3]{x^2} + 2\sqrt[3]x + 4} $$ By multiplying $f(x)$ by $\frac{\sqrt[3]{x^2} + 2\sqrt[3]x + 4}{\sqrt[3]{x^2} + 2\sqrt[3]x + 4}$, show that $f(x) = g(x)$ for all real numbers $x\neq 2$.
The only computation you really need is to see that $$(x^{1/3}-2)*(x^{2/3}+2x^{1/3}+4)=x-8.$$
The rest is simple algebra, for let there be three functions $f_1, f_2, f_3$ such that $f_1*f_3=f_2$, then
$$\frac{f_1}{f_2}*\frac{f_3}{f_3}= \frac{f_2}{f_2*f_3}=\frac{1}{f_3},$$
which proves out result if $f_1=x^{1/3}-2$, $f_2=x-8$ and $f_3=x^{2/3}+2x^{1/3}+4$.