Question 472516

{{{64c^9-27y^6}}} Start with the given expression.



{{{(4c^3)^3-(3y^2)^3}}} Rewrite {{{64c^9}}} as {{{(4c^3)^3}}}. Rewrite {{{27y^6}}} as {{{(3y^2)^3}}}.



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



{{{(4c^3-3y^2)(16c^6+12c^3y^2+9y^4)}}} Multiply


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Answer:

So {{{64c^9-27y^6}}} factors to {{{(4c^3-3y^2)(16c^6+12c^3y^2+9y^4)}}}.


In other words, {{{64c^9-27y^6=(4c^3-3y^2)(16c^6+12c^3y^2+9y^4)}}}