Solve x^3+y^3-3axy=0

$$x ^ { 3 } + y ^ { 3 } - 3 a x y = 0$$

Solve for a

$$\left\{\begin{matrix}a=\frac{x^{3}+y^{3}}{3xy}\text{, }&y\neq 0\text{ and }x\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract x^{3} from both sides. Anything subtracted from zero gives its negation.

$$y^{3}-3axy=-x^{3}$$

Subtract y^{3} from both sides.

$$-3axy=-x^{3}-y^{3}$$

The equation is in standard form.

$$\left(-3xy\right)a=-x^{3}-y^{3}$$

Divide both sides by -3xy.

$$\frac{\left(-3xy\right)a}{-3xy}=\frac{-x^{3}-y^{3}}{-3xy}$$

Dividing by -3xy undoes the multiplication by -3xy.

$$a=\frac{-x^{3}-y^{3}}{-3xy}$$

Divide -x^{3}-y^{3} by -3xy.

$$a=\frac{x^{2}}{3y}+\frac{y^{2}}{3x}$$