Solve left(x-aright)left(x-aright)-left(x+aright)left(x+aright)=aleft(2x-abright)

$$\left( x-a \right) \left( x-a \right) - \left( x+a \right) \left( x+a \right) = a \left( 2x-ab \right)$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=\frac{6x}{a}\text{, }&a\neq 0\\b\in \mathrm{C}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply x-a and x-a to get \left(x-a\right)^{2}.

$$\left(x-a\right)^{2}-\left(x+a\right)\left(x+a\right)=a\left(2x-ab\right)$$

Multiply x+a and x+a to get \left(x+a\right)^{2}.

$$\left(x-a\right)^{2}-\left(x+a\right)^{2}=a\left(2x-ab\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(x-a\right)^{2}.

$$x^{2}-2xa+a^{2}-\left(x+a\right)^{2}=a\left(2x-ab\right)$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x+a\right)^{2}.

$$x^{2}-2xa+a^{2}-\left(x^{2}+2xa+a^{2}\right)=a\left(2x-ab\right)$$

To find the opposite of x^{2}+2xa+a^{2}, find the opposite of each term.

$$x^{2}-2xa+a^{2}-x^{2}-2xa-a^{2}=a\left(2x-ab\right)$$

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

$$-2xa+a^{2}-2xa-a^{2}=a\left(2x-ab\right)$$

Combine -2xa and -2xa to get -4xa.

$$-4xa+a^{2}-a^{2}=a\left(2x-ab\right)$$

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

$$-4xa=a\left(2x-ab\right)$$

Use the distributive property to multiply a by 2x-ab.

$$-4xa=2ax-ba^{2}$$

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

$$2ax-ba^{2}=-4xa$$

Subtract 2ax from both sides.

$$-ba^{2}=-4xa-2ax$$

Combine -4xa and -2ax to get -6xa.

$$-ba^{2}=-6xa$$

The equation is in standard form.

$$\left(-a^{2}\right)b=-6ax$$

Divide both sides by -a^{2}.

$$\frac{\left(-a^{2}\right)b}{-a^{2}}=-\frac{6ax}{-a^{2}}$$

Dividing by -a^{2} undoes the multiplication by -a^{2}.

$$b=-\frac{6ax}{-a^{2}}$$

Divide -6xa by -a^{2}.

$$b=\frac{6x}{a}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{6x}{a}\text{, }&a\neq 0\\b\in \mathrm{R}\text{, }&a=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply x-a and x-a to get \left(x-a\right)^{2}.

$$\left(x-a\right)^{2}-\left(x+a\right)\left(x+a\right)=a\left(2x-ab\right)$$

Multiply x+a and x+a to get \left(x+a\right)^{2}.

$$\left(x-a\right)^{2}-\left(x+a\right)^{2}=a\left(2x-ab\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(x-a\right)^{2}.

$$x^{2}-2xa+a^{2}-\left(x+a\right)^{2}=a\left(2x-ab\right)$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x+a\right)^{2}.

$$x^{2}-2xa+a^{2}-\left(x^{2}+2xa+a^{2}\right)=a\left(2x-ab\right)$$

To find the opposite of x^{2}+2xa+a^{2}, find the opposite of each term.

$$x^{2}-2xa+a^{2}-x^{2}-2xa-a^{2}=a\left(2x-ab\right)$$

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

$$-2xa+a^{2}-2xa-a^{2}=a\left(2x-ab\right)$$

Combine -2xa and -2xa to get -4xa.

$$-4xa+a^{2}-a^{2}=a\left(2x-ab\right)$$

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

$$-4xa=a\left(2x-ab\right)$$

Use the distributive property to multiply a by 2x-ab.

$$-4xa=2ax-ba^{2}$$

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

$$2ax-ba^{2}=-4xa$$

Subtract 2ax from both sides.

$$-ba^{2}=-4xa-2ax$$

Combine -4xa and -2ax to get -6xa.

$$-ba^{2}=-6xa$$

The equation is in standard form.

$$\left(-a^{2}\right)b=-6ax$$

Divide both sides by -a^{2}.

$$\frac{\left(-a^{2}\right)b}{-a^{2}}=-\frac{6ax}{-a^{2}}$$

Dividing by -a^{2} undoes the multiplication by -a^{2}.

$$b=-\frac{6ax}{-a^{2}}$$

Divide -6xa by -a^{2}.

$$b=\frac{6x}{a}$$

Solve for a (complex solution)

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a=\frac{6x}{b}\text{, }&b\neq 0\\a\in \mathrm{C}\text{, }&x=0\text{ and }b=0\end{matrix}\right.$$

Solve for a

$$\left\{\begin{matrix}\\a=0\text{, }&\text{unconditionally}\\a=\frac{6x}{b}\text{, }&b\neq 0\\a\in \mathrm{R}\text{, }&x=0\text{ and }b=0\end{matrix}\right.$$