How do you factor (r+6)^3-216(r+6)3−216?
1 Answer
Explanation:
The difference of cubes identity can be written:
a^3-b^3=(a-b)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
We can use this with
(r+6)^3-216(r+6)3−216
=(r+6)^3-6^3=(r+6)3−63
=((r+6)-6)((r+6)^2+(r+6)(6)+6^2)=((r+6)−6)((r+6)2+(r+6)(6)+62)
=r((r^2+12r+36)+(6r+36)+36)=r((r2+12r+36)+(6r+36)+36)
=r(r^2+18r+108)=r(r2+18r+108)
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+108r2+18r+108
=(r+9)^2-81+108=(r+9)2−81+108
=(r+9)^2+27=(r+9)2+27
=(r+9)^2-(3sqrt(3)i)^2=(r+9)2−(3√3i)2
=((r+9)-3sqrt(3)i)((r+9)+3sqrt(3)i)=((r+9)−3√3i)((r+9)+3√3i)
=(r+9-3sqrt(3)i)(r+9+3sqrt(3)i)=(r+9−3√3i)(r+9+3√3i)
