How do you factor by grouping $m^{3} + 3 m^{2} n + 3 m n^{2} + n^{3}$ $m^3 + 3m^2n + 3mn^2 +n^3$ ?

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Answer 1 · 2015-05-14T16:26:38

$m^{3} + n^{3} + 3 m n (m + n)$ $m^3+n^3 +3mn (m+n)$

= $(m + n) (m^{2} - m n + n^{2}) + 3 m n (m + n)$ $(m+n)(m^2 -mn +n^2)+3mn(m+n)$

= $(m + n) (m^{2} - m n + n^{2} + 3 m n)$ $(m+n)(m^2-mn+n^2 +3mn)$

= $(m + n) (m^{2} + 2 m n + n^{2})$ $(m+n)(m^2+2mn+n^2)$

= $(m + n) {(m + n)}^{2}$ $(m+n)(m+n)^2$

= ${(m + n)}^{3}$ $(m+n)^3$

How do you factor by grouping m3+3m2n+3mn2+n3m^3 + 3m^2n + 3mn^2 +n^3?