Example :  Solve \({x^4}-{\rm{ }}9{x^2} + {\rm{ }}8{\rm{ }} = {\rm{ }}0\)

Solution: Let \(u = {x^2}\) and substitute

\({u^2}-{\rm{ }}9u + {\rm{ }}8{\rm{ }} = {\rm{ }}0\)

Factor this quadratic to get

\((u - 8) (u -1) = 0\)

Solutions for \(“u”\) are 8 and 1.

\({x^2} = 8\) \( \Rightarrow x =  \pm \)\(2\sqrt 2 \)

\({x^2} = {\rm{ }}1 \Rightarrow x =  \pm 1\)

Example: \({\left( {{y^2}-{\rm{ }}2y} \right)^2}-{\rm{ }}11\left( {{y^2}-{\rm{ }}2y} \right){\rm{ }} + {\rm{ }}24{\rm{ }} = 0\)

Solution: Suppose \({y^2}-{\rm{ }}2y{\rm{ }} = {\rm{ }}u\), then the equation becomes 

\({u^2}-{\rm{ }}11u{\rm{ }} + {\rm{ }}24{\rm{ }} = {\rm{ }}0\)

\((u - 8)(u - 3) = 0\) or \(u =8\) and 3

Now \({y^2}-{\rm{ }}2y\; = {\rm{ }}3\) or \(( y - 3)( y + 1) = 0\) or \(y = 3, -1\).

Now \({y^2}-{\rm{ }}2y\; = {\rm{ }}8\) or \(( y - 4)( y + 2) = 0\) or \(y = 4, -2\).

Questions 01: \({\left( {{y^2}-{\rm{ }}2y} \right)^2}-{\rm{ }}11\left( {{y^2}-{\rm{ }}2y} \right){\rm{ }} + {\rm{ }}24{\rm{ }} = 0\)