Solve E(x-m)^2(x+m)-(2x-m)^2=m^2(m-1)+x^3

$$E ( x - m ) ^ { 2 } ( x + m ) - ( 2 x - m ) ^ { 2 } = m ^ { 2 } ( m - 1 ) + x ^ { 3 }$$

Solve for E

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

Steps for Solving Linear Equation

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

$$E\left(x^{2}-2xm+m^{2}\right)\left(x+m\right)-\left(2x-m\right)^{2}=m^{2}\left(m-1\right)+x^{3}$$

Use the distributive property to multiply E by x^{2}-2xm+m^{2}.

$$\left(Ex^{2}-2Exm+Em^{2}\right)\left(x+m\right)-\left(2x-m\right)^{2}=m^{2}\left(m-1\right)+x^{3}$$

Use the distributive property to multiply Ex^{2}-2Exm+Em^{2} by x+m and combine like terms.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}-\left(2x-m\right)^{2}=m^{2}\left(m-1\right)+x^{3}$$

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

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}-\left(4x^{2}-4xm+m^{2}\right)=m^{2}\left(m-1\right)+x^{3}$$

To find the opposite of 4x^{2}-4xm+m^{2}, find the opposite of each term.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}-4x^{2}+4xm-m^{2}=m^{2}\left(m-1\right)+x^{3}$$

Use the distributive property to multiply m^{2} by m-1.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}-4x^{2}+4xm-m^{2}=m^{3}-m^{2}+x^{3}$$

Add 4x^{2} to both sides.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}+4xm-m^{2}=m^{3}-m^{2}+x^{3}+4x^{2}$$

Subtract 4xm from both sides.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}-m^{2}=m^{3}-m^{2}+x^{3}+4x^{2}-4xm$$

Add m^{2} to both sides.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}=m^{3}-m^{2}+x^{3}+4x^{2}-4xm+m^{2}$$

Combine -m^{2} and m^{2} to get 0.

$$Ex^{3}-Ex^{2}m-Exm^{2}+Em^{3}=m^{3}+x^{3}+4x^{2}-4xm$$

Combine all terms containing E.

$$\left(x^{3}-x^{2}m-xm^{2}+m^{3}\right)E=m^{3}+x^{3}+4x^{2}-4xm$$

The equation is in standard form.

$$\left(x^{3}-mx^{2}-xm^{2}+m^{3}\right)E=x^{3}+4x^{2}-4mx+m^{3}$$

Divide both sides by x^{3}-mx^{2}-xm^{2}+m^{3}.

$$\frac{\left(x^{3}-mx^{2}-xm^{2}+m^{3}\right)E}{x^{3}-mx^{2}-xm^{2}+m^{3}}=\frac{x^{3}+4x^{2}-4mx+m^{3}}{x^{3}-mx^{2}-xm^{2}+m^{3}}$$

Dividing by x^{3}-mx^{2}-xm^{2}+m^{3} undoes the multiplication by x^{3}-mx^{2}-xm^{2}+m^{3}.

$$E=\frac{x^{3}+4x^{2}-4mx+m^{3}}{x^{3}-mx^{2}-xm^{2}+m^{3}}$$

Divide 4x^{2}-4xm+m^{3}+x^{3} by x^{3}-mx^{2}-xm^{2}+m^{3}.

$$E=\frac{x^{3}+4x^{2}-4mx+m^{3}}{\left(x-m\right)\left(x^{2}-m^{2}\right)}$$