Let $\begin{array}{r}U=\left\{x\mid x^5-6x^4+11x^3-6x^2=0\right\},A=\left\{x\mid x^2-5x+6=0\right\}, B=\left\{x\mid x^2-3x+2=0\right\}\end{array}$What is $(A\cap B)\text{'}$ equal to ?

$U=\left\{x:x^5-6x^4+11x^3-6x^2=0\right\}$

Solving for values of $x$, we get $\mathrm{U}=\{0,1,2,3\}$

Solving for values of $x$, we get $\mathrm{A}=\left\{\mathrm{x}:{\mathrm{x}}^2-5\mathrm{x}+6=0\right\}$ $\Rightarrow \mathrm{A}=\{2,3\}$

and B $=\left\{x:x^2-3x+2=0\right\}$ Solving for values of x, we get$\Rightarrow \mathrm{B}=\{2,1\}$

  $\begin{array}{l}\mathrm{A}\cap \mathrm{B}=\left\{2\right\}\\ \therefore (\mathrm{A}\cap \mathrm{B}{)}^{\mathrm{\prime }}=\mathrm{U}-(\mathrm{A}\cap \mathrm{B})=\{0,1,2,3\}-\{2\}=\{0,1,3\}\end{array}$