Question 158096
Let {{{y=x+2}}} and {{{z=x-5}}}



So {{{(x+2)^3+(x-5)^3}}} becomes {{{y^3+z^3}}}



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



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



{{{(y+z)((y)^2-(y)(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)}}}



{{{(y+z)(y^2-yz+z^2)}}} Multiply



{{{((x+2)+(x-5))((x+2)^2-(x+2)(x-5)+(x-5)^2)}}} Plug in {{{y=x+2}}} and {{{z=x-5}}}



{{{((x+2)+(x-5))(x^2+4x+4-(x^2-3x-10)+x^2-10x+25)}}} FOIL and expand



{{{((x+2)+(x-5))(x^2+4x+4-x^2+3x+10+x^2-10x+25)}}} Distribute the negative



{{{(2x-3)(x^2-3x+39)}}} Combine like terms.



So {{{(x+2)^3+(x-5)^3}}} factors to {{{(2x-3)(x^2-3x+39)}}}



In other words, {{{(x+2)^3+(x-5)^3=(2x-3)(x^2-3x+39)}}}