Question 537712


{{{x^3-8}}} Start with the given expression.



{{{(x)^3-(2)^3}}} Rewrite {{{x^3}}} as {{{(x)^3}}}. Rewrite {{{8}}} as {{{(2)^3}}}.



{{{(x-2)((x)^2+(x)(2)+(2)^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-2)(x^2+2x+4)}}} Multiply


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Answer:

So {{{x^3-8}}} factors to {{{(x-2)(x^2+2x+4)}}}.


In other words, {{{x^3-8=(x-2)(x^2+2x+4)}}}



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