How do you factor u^3 +v^3 +w^3 −3uvw = (u+v+w)((u+v+w)^2 −3(uv+vw+wu))?

#u^3 +v^3 +w^3 −3uvw = (u+v+w)( (u+v+w)^2 −3(uv+vw+wu))#

1 Answer
Shwetank Mauria
Jan 1, 2018

Please see below.

Explanation:

#u^3+v^3+w^3-3uvw#

= #ul(u^3+v^3color(blue)(+3u^2v+3uv^2))+w^3-ul(3uvwcolor(red)(-3u^2v-3uv^2))#

= #ul((u+v)^3+w^3)-3uv(u+v+w)#

= #(u+v+w)((u+v)^2+w^2-(u+v)w)-3uv(u+v+w)#

Here we have factorized #(u+v)^3+w^3# using #x^3+y^3=(x+y)(x^2+y^2-xy)#, where #x=(u+v)# and #y=w#. Observe that now we can take #u+v+w# ascommon and we get

#(u+v+w)(u^2+2uv+v^2+w^2-uv-uw-3uv)#

= #(u+v+w)(u^2+v^2+w^2-uv-vw-uw)#

These are standard factors of #u^3+v^3+w^3-3uvw#. Now also observe that

#(u+v+w)^2=u^2+v^2+w^2+2uv+2vw+2uw#

hence #u^2+v^2+w^2-uv-vw-uw=(u+v+w)^2-3uv-3vw-3uw#

= #(u+v+w)^2-3(uv+vw+uw)#

and hence #u^3+v^3+w^3-3uvw#

= #(u+v+w)((u+v+w)^2-3(uv+vw+uw))#

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