Question 167557
I'm assuming that you want to factor 



Let {{{w=x-2y}}} and {{{z=x+2y}}}



So the expression {{{(x-2y)^3 + (x+2y)^3}}} then becomes {{{w^3 + z^3}}} (through a simple substitution)



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{{{w^3+z^3}}} Start with the given expression.



{{{(w)^3+(z)^3}}} Rewrite {{{w^3}}} as {{{(w)^3}}}. Rewrite {{{z^3}}} as {{{(z)^3}}}.



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



{{{(w+z)(w^2-wz+z^2)}}} Multiply



{{{((x-2y)+(x+2y))((x-2y)^2-(x-2y)(x+2y)+(x+2y)^2)}}} Plug in {{{w=x-2y}}} and {{{z=x+2y}}}



{{{((x-2y)+(x+2y))((x^2-4xy+4y^2)-(x^2-4y^2)+(x^2+4xy+4y^2))}}} FOIL



{{{(x-2y+x+2y)(x^2-4xy+4y^2-x^2+4y^2+x^2+4xy+4y^2)}}} Distribute



{{{2x(x^2+12y^2)}}} Combine like terms.




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Answer:


So {{{(x-2y)^3 + (x+2y)^3}}} factors to {{{2x(x^2+12y^2)}}}.