Question 200130

{{{2x^3-54y^3}}} Start with the given expression



{{{2(x^3-27y^3)}}} Factor out the GCF {{{2}}}



Now let's focus on the inner expression {{{x^3-27y^3}}}


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{{{x^3-27y^3}}} Start with the given expression.



{{{(x)^3-(3y)^3}}} Rewrite {{{x^3}}} as {{{(x)^3}}}. Rewrite {{{27y^3}}} as {{{(3y)^3}}}.



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




{{{(x-3y)(x^2+3xy+9y^2)}}} Multiply



So {{{x^3-27y^3}}} factors to {{{(x-3y)(x^2+3xy+9y^2)}}}.


In other words, {{{x^3-27y^3=(x-3y)(x^2+3xy+9y^2)}}}



So {{{2(x^3-27y^3)}}} factors further to {{{2(x-3y)(x^2+3xy+9y^2)}}}



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Answer:



So {{{2x^3-54y^3}}} completely factors to {{{2(x-3y)(x^2+3xy+9y^2)}}}

    
    
In other words, {{{2x^3-54y^3=2*(x-3y)*(x^2+3xy+9y^2)}}}