Solve x^2-2xb+b^2=left(x^2-bright)^2

$${ x }^{ 2 } -2xb+ { b }^{ 2 } = { \left( { x }^{ 2 } -b \right) }^{ 2 }$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=\frac{x\left(x+1\right)}{2}\text{, }&\text{unconditionally}\\b\in \mathrm{C}\text{, }&x=1\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}-2xb+b^{2}=\left(x^{2}\right)^{2}-2x^{2}b+b^{2}$$

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

$$x^{2}-2xb+b^{2}=x^{4}-2x^{2}b+b^{2}$$

Add 2x^{2}b to both sides.

$$x^{2}-2xb+b^{2}+2x^{2}b=x^{4}+b^{2}$$

Subtract b^{2} from both sides.

$$x^{2}-2xb+b^{2}+2x^{2}b-b^{2}=x^{4}$$

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

$$x^{2}-2xb+2x^{2}b=x^{4}$$

Subtract x^{2} from both sides.

$$-2xb+2x^{2}b=x^{4}-x^{2}$$

Combine all terms containing b.

$$\left(-2x+2x^{2}\right)b=x^{4}-x^{2}$$

The equation is in standard form.

$$\left(2x^{2}-2x\right)b=x^{4}-x^{2}$$

Divide both sides by -2x+2x^{2}.

$$\frac{\left(2x^{2}-2x\right)b}{2x^{2}-2x}=\frac{x^{4}-x^{2}}{2x^{2}-2x}$$

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

$$b=\frac{x^{4}-x^{2}}{2x^{2}-2x}$$

Divide -x^{2}+x^{4} by -2x+2x^{2}.

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

Solve for b

$$\left\{\begin{matrix}\\b=\frac{x\left(x+1\right)}{2}\text{, }&\text{unconditionally}\\b\in \mathrm{R}\text{, }&x=1\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}-2xb+b^{2}=\left(x^{2}\right)^{2}-2x^{2}b+b^{2}$$

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

$$x^{2}-2xb+b^{2}=x^{4}-2x^{2}b+b^{2}$$

Add 2x^{2}b to both sides.

$$x^{2}-2xb+b^{2}+2x^{2}b=x^{4}+b^{2}$$

Subtract b^{2} from both sides.

$$x^{2}-2xb+b^{2}+2x^{2}b-b^{2}=x^{4}$$

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

$$x^{2}-2xb+2x^{2}b=x^{4}$$

Subtract x^{2} from both sides.

$$-2xb+2x^{2}b=x^{4}-x^{2}$$

Combine all terms containing b.

$$\left(-2x+2x^{2}\right)b=x^{4}-x^{2}$$

The equation is in standard form.

$$\left(2x^{2}-2x\right)b=x^{4}-x^{2}$$

Divide both sides by -2x+2x^{2}.

$$\frac{\left(2x^{2}-2x\right)b}{2x^{2}-2x}=\frac{x^{4}-x^{2}}{2x^{2}-2x}$$

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

$$b=\frac{x^{4}-x^{2}}{2x^{2}-2x}$$

Divide -x^{2}+x^{4} by -2x+2x^{2}.

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

Solve for x (complex solution)

$$x=1$$

$$x=0$$

$$x=\frac{\sqrt{8b+1}-1}{2}$$

$$x=\frac{-\sqrt{8b+1}-1}{2}$$

Solve for x

$$\left\{\begin{matrix}\\x=1\text{; }x=0\text{, }&\text{unconditionally}\\x=\frac{-\sqrt{8b+1}-1}{2}\text{; }x=\frac{\sqrt{8b+1}-1}{2}\text{, }&b\geq -\frac{1}{8}\end{matrix}\right.$$