Question 586082
Here's how you factor {{{n^3+216}}}



{{{n^3+216}}} Start with the given expression.



{{{(n)^3+(6)^3}}} Rewrite {{{n^3}}} as {{{(n)^3}}}. Rewrite {{{216}}} as {{{(6)^3}}}.



{{{(n+6)((n)^2-(n)(6)+(6)^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)}}}



{{{(n+6)(n^2-6n+36)}}} Multiply



So {{{n^3+216}}} factors to {{{(n+6)(n^2-6n+36)}}}.


In other words, {{{n^3+216=(n+6)(n^2-6n+36)}}}




This means that {{{n^5+216n^2}}} completely factors to {{{n^2(n+6)(n^2-6n+36)}}}

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