<s>[INST] 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
 [/INST] When you cube it, you get:
$$
 60 > \left(2+\sqrt[3]{7} \right)^3 =15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
Continue to get
$$
 \frac{60-15}{6\sqrt[3]{7}} > \left(2+\sqrt[3]{7} \right).
$$
Now cube again to get:
$$
 \left(\frac{60-15}{6\sqrt[3]{7}}\right)^3 > 15+6\sqrt[3]{7}\left(2+\sqrt[3]{7} \right).
$$
The rhs is now larger than at the beginning, because $\displaystyle\frac{\frac{60-15}{6\sqrt[3]{7}}}{\sqrt[3]{60}}=\frac{60-15}{6\sqrt[3]{7}\sqrt[3]{60}} >1$. 
To show this we use:
$$
45>6\sqrt[3]{7\times 60} \Rightarrow 45^3=91125 >6^3420=216\cdot 420=90720
$$
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