Question 166270
{{{x^6-64y^6}}} Start with the given expression.



{{{(x^3)^2-64y^6}}} Rewrite {{{x^6}}} as {{{(x^3)^2}}}.



{{{(x^3)^2-(8y^3)^2}}} Rewrite {{{64y^6}}} as {{{(8y^3)^2}}}.



Notice how we have a difference of squares. So let's use the difference of squares formula {{{A^2-B^2=(A+B)(A-B)}}} to factor the expression:



{{{(x^3+8y^3)(x^3-8y^3)}}} Factor the expression using the difference of squares.



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Now let's factor {{{x^3+8y^3}}}



{{{x^3+8y^3}}} Start with the given expression.



{{{(x)^3+(2y)^3}}} Rewrite {{{x^3}}} as {{{(x)^3}}}. Rewrite {{{8y^3}}} as {{{(2y)^3}}}.



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



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



So {{{x^3+8y^3}}} factors to {{{(x+2y)(x^2-2xy+4y^2)}}}.



In other words, {{{x^3+8y^3=(x+2y)(x^2-2xy+4y^2)}}}




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Now let's factor {{{x^3-8y^3}}}



{{{x^3-8y^3}}} Start with the given expression.



{{{(x)^3-(2y)^3}}} Rewrite {{{x^3}}} as {{{(x)^3}}}. Rewrite {{{8y^3}}} as {{{(2y)^3}}}.



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



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



So {{{x^3-8y^3}}} factors to {{{(x-2y)(x^2+2xy+4y^2)}}}.



In other words, {{{x^3-8y^3=(x-2y)(x^2+2xy+4y^2)}}}


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Answer:


So {{{x^6-64y^6}}} completely factors to {{{(x+2y)(x-2y)(x^2-2xy+4y^2)(x^2+2xy+4y^2)}}} (I just rearranged the factors a bit)



Note: the order of the factors does not matter