Question 121494

{{{27p^3-1}}} Start with the given expression.



{{{(3p)^3-(1)^3}}} Rewrite {{{27p^3}}} as {{{(3p)^3}}}. Rewrite {{{1}}} as {{{(1)^3}}}.



{{{(3p-1)((3p)^2+(3p)(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)}}}



{{{(3p-1)(9p^2+3p+1)}}} Multiply


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Answer:

So {{{27p^3-1}}} factors to {{{(3p-1)(9p^2+3p+1)}}}.


In other words, {{{27p^3-1=(3p-1)(9p^2+3p+1)}}}