Question 557686
You are correct, {{{a^3-125}}} is a difference of cubes since {{{a^3}}} is a perfect cube and {{{125=5^3}}} is also a perfect cube.



So {{{a^3-125}}} becomes {{{a^3-5^3}}} and we can use the difference of cubes formula {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}} to get



{{{a^3-5^3=(a-5)(a^2+a*5+5^2)}}}



which simplifies to 



{{{a^3-5^3=(a-5)(a^2+5a+25)}}}



So {{{a^3-125=(a-5)(a^2+5a+25)}}} which means that {{{4a^5-500a^2}}} completely factors to {{{4a^2(a-5)(a^2+5a+25)}}}



In other words, {{{4a^5-500a^2=4a^2(a-5)(a^2+5a+25)}}} for all values of 'a'.



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