Question 177257
{{{x^6y^6-1x^3y^3}}} Start with the given expression



{{{x^3y^3(x^3y^3-1)}}} Factor out the GCF {{{x^3y^3}}}



Now let's focus on the inner expression {{{x^3y^3-1}}}





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{{{x^3y^3-1}}} Start with the inner expression.



{{{(xy)^3-(1)^3}}} Rewrite {{{x^3y^3}}} as {{{(xy)^3}}}. Rewrite {{{1}}} as {{{(1)^3}}}.



{{{(xy-1)((xy)^2+(xy)(1)+(1)^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)}}}



{{{(xy-1)(x^2y^2+xy+1)}}} Multiply


So {{{x^3y^3-1}}} factors to {{{(xy-1)(x^2y^2+xy+1)}}}.




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Answer:

So {{{x^6y^6-x^3y^3}}} completely factors to {{{x^3y^3(xy-1)(x^2y^2+xy+1)}}}