Solve 0=(x-3)^2(x-z)

$$0 = ( x - 3 ) ^ { 2 } ( x - z )$$

Solve for z

$$\left\{\begin{matrix}\\z=x\text{, }&\text{unconditionally}\\z\in \mathrm{R}\text{, }&x=3\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$0=\left(x^{2}-6x+9\right)\left(x-z\right)$$

Use the distributive property to multiply x^{2}-6x+9 by x-z.

$$0=x^{3}-x^{2}z-6x^{2}+6xz+9x-9z$$

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

$$x^{3}-x^{2}z-6x^{2}+6xz+9x-9z=0$$

Subtract x^{3} from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}z-6x^{2}+6xz+9x-9z=-x^{3}$$

Add 6x^{2} to both sides.

$$-x^{2}z+6xz+9x-9z=-x^{3}+6x^{2}$$

Subtract 9x from both sides.

$$-x^{2}z+6xz-9z=-x^{3}+6x^{2}-9x$$

Combine all terms containing z.

$$\left(-x^{2}+6x-9\right)z=-x^{3}+6x^{2}-9x$$

Divide both sides by -x^{2}+6x-9.

$$\frac{\left(-x^{2}+6x-9\right)z}{-x^{2}+6x-9}=-\frac{x\left(3-x\right)^{2}}{-x^{2}+6x-9}$$

Dividing by -x^{2}+6x-9 undoes the multiplication by -x^{2}+6x-9.

$$z=-\frac{x\left(3-x\right)^{2}}{-x^{2}+6x-9}$$

Divide -\left(3-x\right)^{2}x by -x^{2}+6x-9.

$$z=x$$

Solve for x

$$x=z$$

$$x=3$$