Solve left(x+1right)^2-(x+1)(x-1)=0

$${ \left(x+1 \right) }^{ 2 } -(x+1)(x-1)=0$$

Solve for x

$$x=-1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+2x+1-\left(x+1\right)\left(x-1\right)=0$$

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}+2x+1-\left(x^{2}-1\right)=0$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+2x+1-x^{2}+1=0$$

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

$$2x+1+1=0$$

Add 1 and 1 to get 2.

$$2x+2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$2x=-2$$

Divide both sides by 2.

$$x=\frac{-2}{2}$$

Divide -2 by 2 to get -1.

$$x=-1$$