How do you factor #(r+6)^3-216#?

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Answer 1 · 2016-05-02T07:53:41

#(r+6)^3-216=r(r^2+18r+108)#

The difference of cubes identity can be written:

#a^3-b^3=(a-b)(a^2+ab+b^2)#

We can use this with #a=(r+6)# and #b=6# as follows:

#(r+6)^3-216#

#=(r+6)^3-6^3#

#=((r+6)-6)((r+6)^2+(r+6)(6)+6^2)#

#=r((r^2+12r+36)+(6r+36)+36)#

#=r(r^2+18r+108)#

#color(white)()#
Footnote

The remaining quadratic factor cannot be factorised further using Real coefficients.

It can be factorised by completing the square using Complex coefficients:

#r^2+18r+108#

#=(r+9)^2-81+108#

#=(r+9)^2+27#

#=(r+9)^2-(3sqrt(3)i)^2#

#=((r+9)-3sqrt(3)i)((r+9)+3sqrt(3)i)#

#=(r+9-3sqrt(3)i)(r+9+3sqrt(3)i)#