Question 186370


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



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



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



{{{(y-2z)(y^2+2yz+4z^2)}}} Multiply


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Answer:


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



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