Solve dx=sqrt[3]{t(t^2+4)}

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

Solve for d

$$\left\{\begin{matrix}d=\frac{\sqrt[3]{t\left(t^{2}+4\right)}}{x}\text{, }&x\neq 0\\d\in \mathrm{R}\text{, }&t=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply t by t^{2}+4.

$$dx=\sqrt[3]{t^{3}+4t}$$

The equation is in standard form.

$$xd=\sqrt[3]{t^{3}+4t}$$

Divide both sides by x.

$$\frac{xd}{x}=\frac{\sqrt[3]{t\left(t^{2}+4\right)}}{x}$$

Dividing by x undoes the multiplication by x.

$$d=\frac{\sqrt[3]{t\left(t^{2}+4\right)}}{x}$$

Solve for t

$$t=\frac{\sqrt[3]{3}\times 2^{\frac{2}{3}}\left(\sqrt[3]{\sqrt{3\left(27\left(dx\right)^{6}+256\right)}+9\left(dx\right)^{3}}+\sqrt[3]{-\sqrt{3\left(27\left(dx\right)^{6}+256\right)}+9\left(dx\right)^{3}}\right)}{6}$$