<s>[INST] Evaluating $\sum^{7}_{k=1}\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}$ How do I evaluate this?
$$\sum^{7}_{k=1}\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}$$
I know I'm suppose to use substitutions but didn't work out so far.  Help anyone?
 [/INST] Indeed, make a substitution:
Let $\boxed{x=\sqrt[3]{k}}$ and $\boxed{y=\sqrt[3]{k+1}}$.
$$\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}=\frac{1}{x^2+xy+y^2}=\frac{y-x}{(y-x)(y^2+xy+x^2)}=\frac{y-x}{y^3-x^3}={y-x}{(k+1)-k}=y-x=\sqrt[3]{k+1}-\sqrt[3]{k}$$
Now, we know that the whole expression can be expressed as:
$$\therefore \sum^{7}_{k=1}\frac{1}{\sqrt[3]{k^2}+\sqrt[3]{k(k+1)}+\sqrt[3]{(k+1)^2}}=(\sqrt[3]{2}-\sqrt[3]{1})+(\sqrt[3]{3}-\sqrt[3]{2})+\ldots +(\sqrt[3]{8}-\sqrt[3]{7})=\sqrt[3]{8}-1=\boxed{1}$$
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