​ Help? I'm confused

Uhhh idk man really i diont that is to advanced for rn ok

Not sure what you want to do......here's the graph : https://www.desmos.com/calculator/0lpspyn1ud

$f(x)=\sqrt[3]{{x}^{2}+4x}$

${f(x)}^{3}={\sqrt[3]{{x}^{2}+4x}}^{3}$

${f(x)}^{3}={x}^{2}+4x$

${f(x)}^{3}=x(x+4)$

$\frac{{f(x)}^{3}}{x}=\frac{x(x+4)}{x}$

$\frac{{f(x)}^{3}}{x}=x+4$

$\frac{{f(x)}^{3}}{x}-4=x+4-4$

$\frac{{f(x)}^{3}}{x}-4=x+0$

$\frac{{f(x)}^{3}}{x}-4=x$

$x(\frac{{f(x)}^{3}}{x}-4)=x\times x$

${f(x)}^{3}-4x=x\times x$

${f(x)}^{3}-4x={x}^{2}$

${f(x)}^{3}-4x-{x}^{2}={x}^{2}-{x}^{2}$

${f(x)}^{3}-4x-{x}^{2}={0x}^{2}$

${f(x)}^{3}-4x-{x}^{2}=0$

$-{x}^{2}-4x+{f(x)}^{3}=0$

$x = {4 \pm \sqrt{(-4)^2-4(-1)({f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm \sqrt{16-4(-1)({f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm \sqrt{16-(-4)({f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm \sqrt{16+4{f(x)}^{3}} \over 2(-1)}$

$x = {4 \pm \sqrt{4(4+{f(x)}^{3})} \over 2(-1)}$

$x = {4 \pm 2\sqrt{4+{f(x)}^{3}} \over 2(-1)}$

$x = {2(2 \pm 1\sqrt{4+{f(x)}^{3}} )\over 2(-1)}$

$x = {2(2 \pm \sqrt{4+{f(x)}^{3}} )\over 2(-1)}$

$x = {2(2 \pm \sqrt{4+{f(x)}^{3}} )\over -2}$

$x = -1(2 \pm \sqrt{4+{f(x)}^{3}} )$

$x = -(2 \pm \sqrt{4+{f(x)}^{3}} )$

$x = -2 \pm \sqrt{4+{f(x)}^{3}}$