Solve 3x-1-x-1/left(x+1right)^2=3x-1

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

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Variable x cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(x+1\right)^{2}.

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

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

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

Use the distributive property to multiply 3x by x^{2}+2x+1.

$$3x^{3}+6x^{2}+3x+\left(x+1\right)^{2}\left(-1\right)-\left(x-1\right)=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

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

$$3x^{3}+6x^{2}+3x+\left(x^{2}+2x+1\right)\left(-1\right)-\left(x-1\right)=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

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

$$3x^{3}+6x^{2}+3x-x^{2}-2x-1-\left(x-1\right)=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$3x^{3}+5x^{2}+3x-2x-1-\left(x-1\right)=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

Combine 3x and -2x to get x.

$$3x^{3}+5x^{2}+x-1-\left(x-1\right)=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

To find the opposite of x-1, find the opposite of each term.

$$3x^{3}+5x^{2}+x-1-x+1=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

Combine x and -x to get 0.

$$3x^{3}+5x^{2}-1+1=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

Add -1 and 1 to get 0.

$$3x^{3}+5x^{2}=3x\left(x+1\right)^{2}+\left(x+1\right)^{2}\left(-1\right)$$

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

$$3x^{3}+5x^{2}=3x\left(x^{2}+2x+1\right)+\left(x+1\right)^{2}\left(-1\right)$$

Use the distributive property to multiply 3x by x^{2}+2x+1.

$$3x^{3}+5x^{2}=3x^{3}+6x^{2}+3x+\left(x+1\right)^{2}\left(-1\right)$$

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

$$3x^{3}+5x^{2}=3x^{3}+6x^{2}+3x+\left(x^{2}+2x+1\right)\left(-1\right)$$

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

$$3x^{3}+5x^{2}=3x^{3}+6x^{2}+3x-x^{2}-2x-1$$

Combine 6x^{2} and -x^{2} to get 5x^{2}.

$$3x^{3}+5x^{2}=3x^{3}+5x^{2}+3x-2x-1$$

Combine 3x and -2x to get x.

$$3x^{3}+5x^{2}=3x^{3}+5x^{2}+x-1$$

Subtract 3x^{3} from both sides.

$$3x^{3}+5x^{2}-3x^{3}=5x^{2}+x-1$$

Combine 3x^{3} and -3x^{3} to get 0.

$$5x^{2}=5x^{2}+x-1$$

Subtract 5x^{2} from both sides.

$$5x^{2}-5x^{2}=x-1$$

Combine 5x^{2} and -5x^{2} to get 0.

$$0=x-1$$

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

$$x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$x=1$$