What is the standard form of # y(8y + 5)(8y - 5) #?

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Answer 1 · 2016-12-08T23:57:19

#64y^3 - 25y

First, cross multiply the terms in parenthesis:

#y((8y*8y) - (5*8y) + (5*8y) - (5*5))#

#y(64y^2 - 40y + 40y - 25#

#y(64y^2 - 25)#

Then distribute across the terms in parenthesis:

#64y^2y - 25*y#

#64y^3 - 25y#