Question
Factorise:$y^2+\frac{1}{4 y^2}+1-6 y-\frac{3}{y}$
✓
Answer
$y^2+\frac{1}{4 y^2}+1-6 y-\frac{3}{y} $
$=\left(y^2+\frac{1}{4 y^2}+1\right)-\left(6 y+\frac{3}{y}\right) $
$ =\left(y+\frac{1}{2 y}\right)^2-6\left(y+\frac{1}{2 y}\right) $
$=\left(y+\frac{1}{2 y}\right)\left(y+\frac{1}{2 y}-6\right) .$
$=\left(y^2+\frac{1}{4 y^2}+1\right)-\left(6 y+\frac{3}{y}\right) $
$ =\left(y+\frac{1}{2 y}\right)^2-6\left(y+\frac{1}{2 y}\right) $
$=\left(y+\frac{1}{2 y}\right)\left(y+\frac{1}{2 y}-6\right) .$
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