Solve x^2+1/x^2=(x+1/x)^2+y

$$x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = ( x + \frac { 1 } { x } ) ^ { 2 } + y$$

Solve for y

$$y=-2$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x^{2}.

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

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$x^{4}+1=x^{2}\left(x+\frac{1}{x}\right)^{2}+x^{2}y$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x}{x}.

$$x^{4}+1=x^{2}\left(\frac{xx}{x}+\frac{1}{x}\right)^{2}+x^{2}y$$

Since \frac{xx}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

$$x^{4}+1=x^{2}\times \left(\frac{xx+1}{x}\right)^{2}+x^{2}y$$

Do the multiplications in xx+1.

$$x^{4}+1=x^{2}\times \left(\frac{x^{2}+1}{x}\right)^{2}+x^{2}y$$

To raise \frac{x^{2}+1}{x} to a power, raise both numerator and denominator to the power and then divide.

$$x^{4}+1=x^{2}\times \frac{\left(x^{2}+1\right)^{2}}{x^{2}}+x^{2}y$$

Express x^{2}\times \frac{\left(x^{2}+1\right)^{2}}{x^{2}} as a single fraction.

$$x^{4}+1=\frac{x^{2}\left(x^{2}+1\right)^{2}}{x^{2}}+x^{2}y$$

Cancel out x^{2} in both numerator and denominator.

$$x^{4}+1=\left(x^{2}+1\right)^{2}+x^{2}y$$

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

$$x^{4}+1=\left(x^{2}\right)^{2}+2x^{2}+1+x^{2}y$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}+1=x^{4}+2x^{2}+1+x^{2}y$$

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

$$x^{4}+2x^{2}+1+x^{2}y=x^{4}+1$$

Subtract x^{4} from both sides.

$$2x^{2}+1+x^{2}y=x^{4}+1-x^{4}$$

Combine x^{4} and -x^{4} to get 0.

$$2x^{2}+1+x^{2}y=1$$

Subtract 2x^{2} from both sides.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$x^{2}y=-2x^{2}$$

Divide both sides by x^{2}.

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

Dividing by x^{2} undoes the multiplication by x^{2}.

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

Divide -2x^{2} by x^{2}.

$$y=-2$$

Solve for x

$$x\neq 0$$

$$y=-2\text{ and }x\neq 0$$