Solve (x^3+y^3)dx+3xyzdy=0

$$( x ^ { 3 } + y ^ { 3 } ) d x + 3 x y z d y = 0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=-\sqrt[3]{y+3z}y^{\frac{2}{3}}\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply y and y to get y^{2}.

$$\left(x^{3}+y^{3}\right)dx+3xy^{2}zd=0$$

Use the distributive property to multiply x^{3}+y^{3} by d.

$$\left(x^{3}d+y^{3}d\right)x+3xy^{2}zd=0$$

Use the distributive property to multiply x^{3}d+y^{3}d by x.

$$dx^{4}+y^{3}dx+3xy^{2}zd=0$$

Combine all terms containing d.

$$\left(x^{4}+y^{3}x+3xy^{2}z\right)d=0$$

The equation is in standard form.

$$\left(x^{4}+xy^{3}+3xzy^{2}\right)d=0$$

Divide 0 by x^{4}+y^{3}x+3xy^{2}z.

$$d=0$$

Solve for x

$$\left\{\begin{matrix}\\x=-\sqrt[3]{y+3z}y^{\frac{2}{3}}\text{; }x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$