How do you factor #192u^3 - 81#?

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How do you factor #192u^3 - 81#?

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Answer 1 · 2018-03-23T15:09:58

#3(16u^2+12u+9)#

Explanation:

#3(64u^3-27)#
#3[(4u)^3-(3)^3)]#
We know that #a^3-b^3=(a-b)(a^2+ab+b^2)#
In this case, #a=4u# and #b=3#
Therefore:
#(4u)^3-(3)^3=(4u-3)(16u^2+12u+9)#
#3(16u^2+12u+9)# (we still have the 3 leftover from before)

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How do you factor #192u^3 - 81#?

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