Question
Factorise:$p ^2+\frac{1}{ p ^2}-3$
✓
Answer
$p^2+\frac{1}{ p ^2}-3 $
$= P ^2+\frac{1}{ p ^2}-2-1 $
$ =\left( p ^2+\frac{1}{ p ^2}-2 \times p \times \frac{1}{ p }\right)-1$
$=\left( p -\frac{1}{ p }\right)^2-1^2$
$=\left( p -\frac{1}{ p }+1\right)\left( p -\frac{1}{ p }-1\right) .$
$= P ^2+\frac{1}{ p ^2}-2-1 $
$ =\left( p ^2+\frac{1}{ p ^2}-2 \times p \times \frac{1}{ p }\right)-1$
$=\left( p -\frac{1}{ p }\right)^2-1^2$
$=\left( p -\frac{1}{ p }+1\right)\left( p -\frac{1}{ p }-1\right) .$
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