OpenStudy (anonymous):

factor m^3+n^3

4 years ago
OpenStudy (anonymous):

\[m^3+n^3=(m+n)(m^2-mn+n^2)\]

4 years ago
OpenStudy (anonymous):

ok got it,, so will the sigs in the trinomial always alternate the + and the -??

4 years ago
OpenStudy (anonymous):

so th sign in (m <sign> n) is whatever sign is in the m^3 <sign> n^3 and then the first sing in the (m^2 <sign> mn +n^2) is opposite

4 years ago
OpenStudy (anonymous):

i think i am finally getting this tytytytyyt!!!!!

4 years ago
OpenStudy (anonymous):

\[m^3+n^3=(m+n)(m^2-mn+n^2)\\m^3-n^3=(m-n)(m^2+mn+n^2)\]

4 years ago
OpenStudy (anonymous):

no problem :)

4 years ago
OpenStudy (anonymous):

1 more ?? pls,,, x^2 + y^2 factors into (x-y) (x+y) correct??

4 years ago
OpenStudy (anonymous):

\[x^2-y^2=(x+y)(x-y)\] and sometimes this one is useful: \[x^2+y^2=(x+y)^2-2xy\]

4 years ago
OpenStudy (anonymous):

kk tytyty

4 years ago
OpenStudy (anonymous):

no problem :)

4 years ago
OpenStudy (anonymous):

the best of luck to you!

4 years ago