How do you factor #y= x^3 + 9x^2 + 27x + 27# ?

1 Answer
George C.
Dec 2, 2015

#y = x^3+9x^2+27x+27 = (x+3)^3#

Explanation:

Notice that #x^3# is a perfect cube and #27 = 3^3# is also a perfect cube.

So try #(x+3)^3# to see if the middle two terms match.

In general #(a+b)^3 = a^3+3a^2b+3ab^2+b^3#, so:

#(x+3)^3 = (x^3) + 3(x^2)(3)+3(x)(3^2)+(3^3)#

#=x^3+9x^2+27x+27#

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