First, note that n2−2n+1,n2−1,n2+2n+1 all can be factored as:
n2−2n+1=(n−1)2
n2−1=(n−1)(n+1)
n2+2n+1=(n+1)2
Therefore, the given sum
999∑n=413√n2−2n+1+3√n2+2n+1+3√n2−1
is hence is equivalent to
999∑n=41((n−1)2)13+((n+1)2)13+((n−1)(n+1))13
=999∑n=41((n−1)13)2+((n+1)13)2+((n−1)(n+1))13
\displaystyle =\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black}\color{black})}
\displaystyle =\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})}(*)
I will stop momentarily here, because I want you to focus on the last expression above, to see if something has (or will) come to you mind when you begin to think, reflect and recall.
I have faith in you that you are already hit on a great idea, haven't you?
Yeah! That exact expression \displaystyle (\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black}) is the factor that we would get when we factor the difference of cubes:
\displaystyle \color{black}\bbox[5px,orange]{a^3-b^3=(a-b)(a^2+ab+b^2)}
We need to make optimal use of this identity from now on to our problem, that is to say, we need to multiply the top and bottom of the expression in (*) with the quantity a-b, i.e. (n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}} so to obtain:
\displaystyle \sum_{n=4}^{999}\dfrac{1}{\sqrt[3]{n^2-2n+1}+\sqrt[3]{n^2+2n+1}+\sqrt[3]{n^2-1}}
\displaystyle =\sum_{n=4}^{999}\dfrac{1}{((n-1)^\frac{1}{3})^2+((n+1)^\frac{1}{3})^2+((n-1)(n+1))^\frac{1}{3}}
\displaystyle =\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black})^2+(\color{yellow}\bbox[5px,purple]{(n-1)^\frac{1}{3}}\color{yellow}\bbox[5px,green]{(n+1)^\frac{1}{3}}\color{black}\color{black})}
\displaystyle =\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})}(*)
\displaystyle =\color{black}\sum_{n=4}^{999}\dfrac{1}{(\color{yellow}\bbox[5px,purple]{a}\color{black})^2+(\color{yellow}\bbox[5px,green]{b}\color{black})^2+(\color{yellow}\bbox[5px,purple]{a}\color{yellow}\bbox[5px,green]{b}\color{black}\color{black})}\dfrac{\cdot \color{black}(\color{yellow}\bbox[5px,purple]{a}\color{black}-\color{yellow}\bbox[5px,green]{b}\color{black})}{\cdot \color{black}(\color{yellow}\bbox[5px,purple]{a}\color{black}-\color{yellow}\bbox[5px,green]{b}\color{black})}
\displaystyle =\sum_{n=4}^{999} \dfrac{\color{black}(\color{yellow}\bbox[5px,purple]{a}\color{black}-\color{yellow}\bbox[5px,green]{b}\color{black})}{\color{black}(\color{yellow}\bbox[5px,purple]{a^3}\color{black}-\color{yellow}\bbox[5px,green]{b^3}\color{black})}
\displaystyle =\sum_{n=4}^{999}\dfrac{(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}}{((n-1)^{\frac{1}{3}})^3-((n+1)^{\frac{1}{3}})^3}
\displaystyle =\sum_{n=4}^{999}\dfrac{(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}}{n-1-n-1}
\displaystyle =\sum_{n=4}^{999}\dfrac{(n-1)^{\frac{1}{3}}-(n+1)^{\frac{1}{3}}}{-2}
\displaystyle =\sum_{n=4}^{999}\dfrac{(n+1)^{\frac{1}{3}}-(n-1)^{\frac{1}{3}}}{2}
=\dfrac{1}{2}[(5^{\frac{1}{3}}-3^{\frac{1}{3}})+(6^{\frac{1}{3}}-4^{\frac{1}{3}})+(7^{\frac{1}{3}}-5^{\frac{1}{3}})+(8^{\frac{1}{3}}-6^{\frac{1}{3}})+\cdots
\,\,\,\,\,\,\,\,\,\,\,\,+(998^{\frac{1}{3}}-996^{\frac{1}{3}})+(999^{\frac{1}{3}}-997^{\frac{1}{3}})+(1000^{\frac{1}{3}}-998^{\frac{1}{3}})]
=\dfrac{1}{2}[(\cancel{5^{\frac{1}{3}}}-3^{\frac{1}{3}})+(\cancel{6^{\frac{1}{3}}}-4^{\frac{1}{3}})+(\cancel{7^{\frac{1}{3}}}-\cancel{5^{\frac{1}{3}}})+(\cancel{8^{\frac{1}{3}}}-\cancel{6^{\frac{1}{3}}})+\cdots
\,\,\,\,\,\,\,\,\,\,\,\,+(\cancel{998^{\frac{1}{3}}}-\cancel{996^{\frac{1}{3}}})+(999^{\frac{1}{3}}-\cancel{997^{\frac{1}{3}}})+(1000^{\frac{1}{3}}-\cancel{998^{\frac{1}{3}}})]
=\dfrac{1}{2}[-3^{\frac{1}{3}}-4^{\frac{1}{3}}+999^{\frac{1}{3}}+1000^{\frac{1}{3}}]
=\dfrac{1}{2}[-3^{\frac{1}{3}}-4^{\frac{1}{3}}+999^{\frac{1}{3}}+10]
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