Question:

Given \(a = n^{1/3} - n^{-1/3}\) and \(b = n^{1/3}+n^{-1/3}\). Step 1: Evaluate \(b^2 - a^2\) Using the difference of squares identity, \(b^2 - a^2 = (b+a)(b-a)\). First, calculate \(b+a\): $$b+a = (n^{1/3}+n^{-1/3}) + (n^{1/3}-n^{-1/3}) = 2n^{1/3}$$ Next, calculate \(b-a\): $$b-a = (n^{1/3}+n^{-1/3}) - (n^{1/3}-n^{-1/3}) = 2n^{-1/3}$$ Now, multiply these results: $$b^2 - a^2 = (2n^{1/3})(2n^{-1/3}) = 4n^{1/3-1/3} = 4n^0 = 4$$ Step 2: Evaluate \(a^3\) Using the binomial expansion \((x-y)^3 = x^3 - y^3 - 3xy(x-y)\): $$a^3 = (n^{1/3}-n^{-1/3})^3$$ $$a^3 = (n^{1/3})^3 - (n^{-1/3})^3 - 3(n^{1/3})(n^{-1/3})(n^{1/3}-n^{-1/3})$$ $$a^3 = n - n^{-1} - 3(1)(a)$$ $$a^3 = n - \frac{1}{n} - 3a$$ Step 3: Evaluate \(b^3\) Using the binomial expansion \((x+y)^3 = x^3 + y^3 + 3xy(x+y)\): $$b^3 = (n^{1/3}+n^{-1/3})^3$$ $$b^3 = (n^{1/3})^3 + (n^{-1/3})^3 + 3(n^{1/3})(n^{-1/3})(n^{1/3}+n^{-1/3})$$ $$b^3 = n + n^{-1} + 3(1)(b)$$ $$b^3 = n + \frac{1}{n} + 3b$$ Rearranging this expression yields: $$b^3 = 3b + \left(n + \frac{1}{n}\right)$$ Step 4: Evaluate \(ab\) Using the difference of squares identity, \((x-y)(x+y) = x^2 - y^2\): $$ab = (n^{1/3}-n^{-1/3})(n^{1/3}+n^{-1/3})$$ $$ab = (n^{1/3})^2 - (n^{-1/3})^2$$ $$ab = n^{2/3} - n^{-2/3}$$ The following statements hold: 1. \(b^2 - a^2 = 4\) 2. \(b^3 = 3b + (n + 1/n)\) 3. \(ab = n^{2/3} - n^{-2/3}\)