Question 4
If $$x = \sqrt[3]{49}+\sqrt[3]{42}+\sqrt[]{36}$$, then the value of $$x-\frac{1}{x^2}$$ is
Solution
Let $$p=\sqrt[3]{7}$$ and $$q=\sqrt[3]{6}$$. Then $$x=p^2+pq+q^2=\frac{p^3-q^3}{p-q}=\frac{1}{p-q}$$, so $$\frac{1}{x}=p-q$$. Thus $$x-\frac{1}{x^2}=p^2+pq+q^2-(p-q)^2=3pq=3\sqrt[3]{42}$$.
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