Question 384527


{{{64+27w^3}}} Start with the given expression.



{{{(4)^3+(3w)^3}}} Rewrite {{{64}}} as {{{(4)^3}}}. Rewrite {{{27w^3}}} as {{{(3w)^3}}}.



{{{(4+3w)((4)^2-(4)(3w)+(3w)^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)}}}



{{{(4+3w)(16-12w+9w^2)}}} Multiply


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Answer:


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


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



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Jim