Question
Factorise : $\frac{1}{4}(a+b)^2-\frac{9}{16}(2 a-b)^2$
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Answer
$ \frac{1}{4}(a+b)^2-\frac{9}{16}(2 a-b)^2$
$ =\frac{1}{4}\left[(a+b)^2-\frac{9}{4}(2 a-b)^2\right]$
$ =\frac{1}{4}\left[(a+b)^2-\left[\frac{3}{2}(2 a-b)^2\right]\right]$
$ =\frac{1}{4}\left[\left(a+b+\frac{3}{2}(2 a-b)\right)\left(a+b-\frac{3}{2}(2 a-b)\right)\right]$
$ =\frac{1}{4}\left[\left(a+b+3 a-\frac{3 b}{2}\right)\left(a+b-3 a+\frac{3 b}{2}\right)\right]$
$ =\frac{1}{4}\left[\left(4 a-\frac{b}{2}\right)\left(\frac{5 b}{2}-2 a\right)\right]$
$ =\frac{1}{4}\left[\left(\frac{8 a-b}{2}\right)\left(\frac{5 b-4 a}{2}\right)\right]$
$ =\frac{1}{4}\left[\frac{1}{4}(8 a-b)(5 b-4 a)\right]$
$ =\frac{1}{16}(8 a-b)(5 b-4 a)$
$ =\frac{1}{4}\left[(a+b)^2-\frac{9}{4}(2 a-b)^2\right]$
$ =\frac{1}{4}\left[(a+b)^2-\left[\frac{3}{2}(2 a-b)^2\right]\right]$
$ =\frac{1}{4}\left[\left(a+b+\frac{3}{2}(2 a-b)\right)\left(a+b-\frac{3}{2}(2 a-b)\right)\right]$
$ =\frac{1}{4}\left[\left(a+b+3 a-\frac{3 b}{2}\right)\left(a+b-3 a+\frac{3 b}{2}\right)\right]$
$ =\frac{1}{4}\left[\left(4 a-\frac{b}{2}\right)\left(\frac{5 b}{2}-2 a\right)\right]$
$ =\frac{1}{4}\left[\left(\frac{8 a-b}{2}\right)\left(\frac{5 b-4 a}{2}\right)\right]$
$ =\frac{1}{4}\left[\frac{1}{4}(8 a-b)(5 b-4 a)\right]$
$ =\frac{1}{16}(8 a-b)(5 b-4 a)$
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