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Question

Factorise the following: a2b2+c2d2a2c2b2d2

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Solution

Given that: a2b2+c2d2a2c2b2d2
=(a2b2a2c2)+(c2d2b2d2), (group pair of terms which have common)
=a2(b2c2)d2(b2c2) (factor each binomials)
=(b2c2)(a2d2), (factor out common factor (b2c2))
=(b+c)(bc)(a+d)(ad),------(since x2y2=(x+y)(xy))

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