Question 184057


{{{8x^6+125y^3}}} Start with the given expression.



{{{(2x^2)^3+(5y)^3}}} Rewrite {{{8x^6}}} as {{{(2x^2)^3}}}. Rewrite {{{125y^3}}} as {{{(5y)^3}}}.



{{{(2x^2+5y)((2x^2)^2-(2x^2)(5y)+(5y)^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)}}}



{{{(2x^2+5y)(4x^4-10x^2y+25y^2)}}} Multiply


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Answer:


So {{{8x^6+125y^3}}} factors to {{{(2x^2+5y)(4x^4-10x^2y+25y^2)}}}.


In other words, {{{8x^6+125y^3=(2x^2+5y)(4x^4-10x^2y+25y^2)}}}