Question 603541


{{{64a^12-m^3n^12}}} Start with the given expression.



{{{(4a^4)^3-(mn^4)^3}}} Rewrite {{{64a^12}}} as {{{(4a^4)^3}}}. Rewrite {{{m^3n^12}}} as {{{(mn^4)^3}}}.



{{{(4a^4-mn^4)((4a^4)^2+(4a^4)(mn^4)+(mn^4)^2)}}} Now factor by using the difference of cubes formula. Remember the <a href="http://www.purplemath.com/modules/specfact2.htm">difference of cubes formula</a> is {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}}



{{{(4a^4-mn^4)(16a^8+4a^4mn^4+m^2n^8)}}} Multiply


-----------------------------------

Answer:

So {{{64a^12-m^3n^12}}} factors to {{{(4a^4-mn^4)(16a^8+4a^4mn^4+m^2n^8)}}}.


In other words, {{{64a^12-m^3n^12=(4a^4-mn^4)(16a^8+4a^4mn^4+m^2n^8)}}}