Solve (1)(a-b)(a-b-c)+(a+b)(-a+b+c)=-2b

$$( 1 ) ( a - b ) ( a - b - c ) + ( a + b ) ( - a + b + c ) = - 2 b$$

Solve for a

$$\left\{\begin{matrix}\\a=b+c+1\text{, }&\text{unconditionally}\\a\in \mathrm{R}\text{, }&b=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 1 by a-b.

$$\left(a-b\right)\left(a-b-c\right)+\left(a+b\right)\left(-a+b+c\right)=-2b$$

Use the distributive property to multiply a-b by a-b-c and combine like terms.

$$a^{2}-2ab-ac+b^{2}+bc+\left(a+b\right)\left(-a+b+c\right)=-2b$$

Use the distributive property to multiply a+b by -a+b+c.

$$a^{2}-2ab-ac+b^{2}+bc+a\left(-a\right)+ab+ac+b\left(-a\right)+b^{2}+bc=-2b$$

Combine -2ab and ab to get -ab.

$$a^{2}-ab-ac+b^{2}+bc+a\left(-a\right)+ac+b\left(-a\right)+b^{2}+bc=-2b$$

Combine -ac and ac to get 0.

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

Combine b^{2} and b^{2} to get 2b^{2}.

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

Combine bc and bc to get 2bc.

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

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

$$a^{2}-ab+2b^{2}+2bc+a^{2}\left(-1\right)+b\left(-1\right)a=-2b$$

Combine a^{2} and a^{2}\left(-1\right) to get 0.

$$-ab+2b^{2}+2bc+b\left(-1\right)a=-2b$$

Combine -ab and b\left(-1\right)a to get -2ab.

$$-2ab+2b^{2}+2bc=-2b$$

Subtract 2b^{2} from both sides.

$$-2ab+2bc=-2b-2b^{2}$$

Subtract 2bc from both sides.

$$-2ab=-2b-2b^{2}-2bc$$

The equation is in standard form.

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

Divide both sides by -2b.

$$\frac{\left(-2b\right)a}{-2b}=-\frac{2b\left(b+c+1\right)}{-2b}$$

Dividing by -2b undoes the multiplication by -2b.

$$a=-\frac{2b\left(b+c+1\right)}{-2b}$$

Divide -2b\left(1+b+c\right) by -2b.

$$a=b+c+1$$

Solve for b

$$b=a-c-1$$

$$b=0$$