Question 202948


{{{64x^3+27y^3}}} Start with the given expression.



{{{(4x)^3+(3y)^3}}} Rewrite {{{64x^3}}} as {{{(4x)^3}}}. Rewrite {{{27y^3}}} as {{{(3y)^3}}}.



{{{(4x+3y)((4x)^2-(4x)(3y)+(3y)^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)}}}



{{{(4x+3y)(16x^2-12xy+9y^2)}}} Multiply


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Answer:


So {{{64x^3+27y^3}}} factors to {{{(4x+3y)(16x^2-12xy+9y^2)}}}.



In other words, {{{64x^3+27y^3=(4x+3y)(16x^2-12xy+9y^2)}}}