Question 182028


{{{27y^4+8y}}} Start with the given expression



{{{y(27y^3+8)}}} Factor out the GCF {{{y}}}



Now let's focus on the inner expression {{{27y^3+8}}}



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{{{27y^3+8}}} Start with the given expression.



{{{(3y)^3+(2)^3}}} Rewrite {{{27y^3}}} as {{{(3y)^3}}}. Rewrite {{{8}}} as {{{(2)^3}}}.



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



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



So {{{27y^3+8}}} factors to {{{(3y+2)(9y^2-6y+4)}}}.


In other words, {{{27y^3+8=(3y+2)(9y^2-6y+4)}}}


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So this means that {{{y(27y^3+8)}}} factors to {{{y(3y+2)(9y^2-6y+4)}}} 



Answer:

So {{{27y^4+8y}}} completely factors to {{{y(3y+2)(9y^2-6y+4)}}}