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

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

Solve for x

$$x=\frac{1}{2}=0.5$$

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}.

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

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

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

Subtract x^{3} from both sides.

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

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

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

Add 3x^{2} to both sides.

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

Combine -3x^{2} and 3x^{2} to get 0.

$$3x-1=x$$

Subtract x from both sides.

$$3x-1-x=0$$

Combine 3x and -x to get 2x.

$$2x-1=0$$

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

$$2x=1$$

Divide both sides by 2.

$$x=\frac{1}{2}$$