Solve (x-y)^3=A(x+y).text{provethat}(2x+y)dy/dx=x+2y

$$( x - y ) ^ { 3 } = A ( x + y ) . \text { prove that } ( 2 x + y ) \frac { d y } { d x } = x + 2 y$$

Solve for A

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

Steps for Solving Linear Equation

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

$$x^{3}-3x^{2}y+3xy^{2}-y^{3}=A\left(x+y\right)$$

Use the distributive property to multiply A by x+y.

$$x^{3}-3x^{2}y+3xy^{2}-y^{3}=Ax+Ay$$

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

$$Ax+Ay=x^{3}-3x^{2}y+3xy^{2}-y^{3}$$

Combine all terms containing A.

$$\left(x+y\right)A=x^{3}-3x^{2}y+3xy^{2}-y^{3}$$

The equation is in standard form.

$$\left(x+y\right)A=x^{3}+3xy^{2}-y^{3}-3yx^{2}$$

Divide both sides by x+y.

$$\frac{\left(x+y\right)A}{x+y}=\frac{\left(x-y\right)^{3}}{x+y}$$

Dividing by x+y undoes the multiplication by x+y.

$$A=\frac{\left(x-y\right)^{3}}{x+y}$$