Prove $\sqrt[3]{60}>2+\sqrt[3]{7}$ Prove $$\sqrt[3]{60}>2+\sqrt[3]{7}$$
I try to both sides of the cubic equation, but it is quite complicated

Well, I don't see any alternative way..
$$60\overset{?}> 8+6\sqrt[3]{7^2}+12\sqrt[3]7+7 $$
$$45\overset{?}> 6\sqrt[3]{7^2}+12\sqrt[3]7 $$
$$15\overset{?}> 2\sqrt[3]{7^2}+4\sqrt[3]7 $$
Well, we could raise it to cubic, but that's really not nice.
What about considering the roots of $2x^2+4x-15 =2(x+1)^2-17$, and finally comparing if $\sqrt[3]7$ is between its roots..
$$\sqrt{\frac{17}2} -1 \overset{?}> \sqrt[3]7 $$
A bit nicer perhaps.. taking cubes:
$$\frac{17}{2}\sqrt{\frac{17}{2}}-3\cdot\frac{17}2+3\cdot\sqrt{\frac{17}2}-1 \overset{?}>7
$$
$$23\sqrt{\frac{17}2} \overset{?}> 16+3\cdot 17 = 67$$
and finally this leads to
$$8993 = 23^2\cdot 17 > 67^2\cdot 2 = 8978 $$
