Evaluate:- $\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$

Question

Evaluate:- $\dfrac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$

What I Tried:- Let $a = 4 , b = \sqrt{15} , c = 6, d= \sqrt{35}$ . Then I get :- $$\rightarrow \frac{[a + b]^{3/2} + [a - b]^{3/2}}{[c + d]^{3/2} - [c - d]^{3/2}}$$ Now I can put the formulas $(a^3 + b^3)$ and $(c^3 - d^3)$ $$\rightarrow \frac{[(a + b)^{1/2} + (a - b)^{1/2}][(a + b) - \sqrt{a^2 - b^2} + (a - b)]}{[(c + d)^{1/2} - (c - d)^{1/2}][(c + d) + \sqrt{c^2 - d^2} + (c - d]}$$ $$\rightarrow \frac{7[(a + b)^{1/2} + (a - b)^{1/2}]}{13[(c + d)^{1/2} - (c - d)^{1/2}]}$$ From here, I do not know how to proceed. Can anyone help me?

Answer 1

Simply square the expression. A lot of cancelation will happen and you'll end up in $490/1690$. Now take square root to get the final answer $7/13$.

Answer 2

$$\sqrt{8+2\sqrt{15}}=\sqrt5+\sqrt3$$

Answer 3

$$\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$$ $$=\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}\times\frac{2^{3/2}}{2^{3/2}}$$ $$=\frac{[8+2\sqrt{15}]^{3/2} + [8-2\sqrt{15}]^{3/2}}{[12+2\sqrt{35}]^{3/2} - [12-2\sqrt{35}]^{3/2}}$$ $$=\frac{[\sqrt5+\sqrt3]^3 + [\sqrt5-\sqrt3]^3}{[\sqrt7+\sqrt5]^3 - [\sqrt7-\sqrt5]^3}$$ $$=\frac{28\sqrt{5}}{52\sqrt{5}}$$ $$=\frac{7}{13}$$