Solve 1=A(x-1)^3

$$1 = A ( x - 1 ) ^ { 3 }$$

Solve for A

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

$$x\neq 1$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.

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

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

$$1=Ax^{3}-3Ax^{2}+3Ax-A$$

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

$$Ax^{3}-3Ax^{2}+3Ax-A=1$$

Combine all terms containing A.

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

Divide both sides by x^{3}-3x^{2}+3x-1.

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

Dividing by x^{3}-3x^{2}+3x-1 undoes the multiplication by x^{3}-3x^{2}+3x-1.

$$A=\frac{1}{x^{3}-3x^{2}+3x-1}$$

Divide 1 by x^{3}-3x^{2}+3x-1.

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

Solve for x

$$x=1+\frac{1}{\sqrt[3]{A}}$$

$$A\neq 0$$