Solve (2x^3-4y^3)^2=dy

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

Solve for d (complex solution)

$$\left\{\begin{matrix}d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}\text{, }&y\neq 0\\d\in \mathrm{C}\text{, }&x=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x^{3}-4y^{3}\right)^{2}.

$$4\left(x^{3}\right)^{2}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$4x^{6}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$4x^{6}-16x^{3}y^{3}+16y^{6}=dy$$

Swap sides so that all variable terms are on the left hand side.

$$dy=4x^{6}-16x^{3}y^{3}+16y^{6}$$

The equation is in standard form.

$$yd=4x^{6}-16x^{3}y^{3}+16y^{6}$$

Divide both sides by y.

$$\frac{yd}{y}=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

Dividing by y undoes the multiplication by y.

$$d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

Solve for d

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

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x^{3}-4y^{3}\right)^{2}.

$$4\left(x^{3}\right)^{2}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$4x^{6}-16x^{3}y^{3}+16\left(y^{3}\right)^{2}=dy$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$4x^{6}-16x^{3}y^{3}+16y^{6}=dy$$

Swap sides so that all variable terms are on the left hand side.

$$dy=4x^{6}-16x^{3}y^{3}+16y^{6}$$

The equation is in standard form.

$$yd=4x^{6}-16x^{3}y^{3}+16y^{6}$$

Divide both sides by y.

$$\frac{yd}{y}=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

Dividing by y undoes the multiplication by y.

$$d=\frac{4\left(x^{3}-2y^{3}\right)^{2}}{y}$$

Solve for x (complex solution)

$$x\in \frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}+\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{4\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}+\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{2\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}+\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{4\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}-\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}-\sqrt{d}\right)}}{2},\frac{2^{\frac{2}{3}}e^{\frac{2\pi i}{3}}\sqrt[3]{\sqrt{y}\left(4y^{\frac{5}{2}}-\sqrt{d}\right)}}{2}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{2^{\frac{2}{3}}\sqrt[3]{4y^{3}-\sqrt{dy}}}{2}\text{; }x=\frac{2^{\frac{2}{3}}\sqrt[3]{4y^{3}+\sqrt{dy}}}{2}\text{, }&d\leq 0\text{ and }y\leq 0\\x=\frac{2^{\frac{2}{3}}\sqrt[6]{y}\sqrt[3]{4y^{\frac{5}{2}}-\sqrt{d}}}{2}\text{; }x=\frac{2^{\frac{2}{3}}\sqrt[6]{y}\sqrt[3]{4y^{\frac{5}{2}}+\sqrt{d}}}{2}\text{, }&y\geq 0\text{ and }d\geq 0\end{matrix}\right.$$