Evaluate $( \sqrt{288} + \sqrt{119})^{3/2} - ( \sqrt{288} - \sqrt{119})^{3/2}$

Question

I need to find the value of

$$( \sqrt{288} + \sqrt{119})^{3/2} - ( \sqrt{288} - \sqrt{119})^{3/2}$$

I tried using the identity $a^{3} - b^{3}$ but couldn't reach very far and got stuck.

Answer 1

Let $$I=( \sqrt{288} + \sqrt{119})^{3/2} - ( \sqrt{288} - \sqrt{119})^{3/2}$$ then $$\begin{aligned} I^2 &=2\sqrt{288}^3-2\cdot 288\sqrt{288-119}+6\sqrt{288}\cdot 119+2\cdot 119\sqrt{288-119} \\ &= 2\sqrt{288}^3-4394+714\sqrt{288} \\ &= 15480\sqrt{2}-4394. \end{aligned}$$ Seeing as this can't be "simplified" further, I suspect the "neatest" form is then $$I=\sqrt{15480\sqrt{2}-4394}.$$

Answer 2

Let $$x= (\sqrt{288} + \sqrt{119})^{3/2} - ( \sqrt{288} - \sqrt{119})^{3/2}$$

and $a= (\sqrt{288} + \sqrt{119})^{3/2}$ and $b= ( \sqrt{288} - \sqrt{119})^{3/2}$

then $$a^2+b^2 = 2\sqrt{288}^3 + 6\sqrt{288}\sqrt{119}^2 = 24\sqrt{2}(288+3\cdot 119)$$ $$ab = (288-119)^{3/2}= 13^3 =2197$$

so $$x = \sqrt{(a-b)^2} = \sqrt{15480\sqrt{2}-4394}$$