Question:

Given $a = n^{1/3} - n^{-1/3}$ and $b = n^{1/3}+n^{-1/3}$. Step 1: Evaluate $b^2 - a^2$. $$b^2 - a^2 = (b+a)(b-a)$$ Substitute the expressions for $a$ and $b$: $$b+a = (n^{1/3}+n^{-1/3}) + (n^{1/3}-n^{-1/3}) = 2n^{1/3}$$ $$b-a = (n^{1/3}+n^{-1/3}) - (n^{1/3}-n^{-1/3}) = 2n^{-1/3}$$ Therefore, $$b^2 - a^2 = (2n^{1/3})(2n^{-1/3}) = 4n^{1/3-1/3} = 4n^0 = 4$$ Step 2: Evaluate $a^3$. Using the identity $(x-y)^3 = x^3 - y^3 - 3xy(x-y)$: $$a^3 = (n^{1/3}-n^{-1/3})^3$$ $$a^3 = (n^{1/3})^3 - (n^{-1/3})^3 - 3(n^{1/3})(n^{-1/3})(n^{1/3}-n^{-1/3})$$ $$a^3 = n - n^{-1} - 3(1)(a)$$ $$a^3 = n - \frac{1}{n} - 3a$$ Rearranging the terms, we get: $$a^3 + 3a = n - \frac{1}{n}$$ This can also be written as: $$a^3 = -3a + \left(n - \frac{1}{n}\right)$$ Or, to match the given form: $$a^3 = 3a - \left(n - \frac{1}{n}\right) + 2\left(n - \frac{1}{n}\right)$$ This implies there was a sign error in the original statement. The correct relation is $a^3 = n - \frac{1}{n} - 3a$. Step 3: Evaluate $b^3$. Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$: $$b^3 = (n^{1/3}+n^{-1/3})^3$$ $$b^3 = (n^{1/3})^3 + (n^{-1/3})^3 + 3(n^{1/3})(n^{-1/3})(n^{1/3}+n^{-1/3})$$ $$b^3 = n + n^{-1} + 3(1)(b)$$ $$b^3 = n + \frac{1}{n} + 3b$$ Rearranging the terms, we get: $$b^3 - 3b = n + \frac{1}{n}$$ This can also be written as: $$b^3 = 3b + \left(n + \frac{1}{n}\right)$$ Step 4: Evaluate $ab$. $$ab = (n^{1/3}-n^{-1/3})(n^{1/3}+n^{-1/3})$$ Using the identity $(x-y)(x+y) = x^2 - y^2$: $$ab = (n^{1/3})^2 - (n^{-1/3})^2$$ $$ab = n^{2/3} - n^{-2/3}$$