Nice number

First we have that $x^{2} = \frac{1}{4} (7^{\frac{2}{3}} - 2 + 7^{-\frac{2}{3}})$ .

So $1 + x^{2} = \frac{1}{4} (7^{\frac{2}{3}} + 2 + 7^{-\frac{2}{3}}) = [\frac{1}{2} (7^{\frac{1}{3}} + 7^{-\frac{1}{3}})]^{2}$ .

Thus $x + \sqrt{1 + x^{2}} = \frac{1}{2} (7^{\frac{1}{3}} - 7^{-\frac{1}{3}}) + \frac{1}{2} (7^{\frac{1}{3}} + 7^{-\frac{1}{3}}) = 7^{\frac{1}{3}}$ ,

and so $(x + \sqrt{1 + x^{2}})^{3} = \boxed{7}$ .

Let cube root of 7 be "a". Now, solve it.

let 7^(1/3) be y. x²=(a^4 -2a²+1)/4a² 1+x² = (a^4 +2a²+1)/4a²=((a²+1)/2a)² (1+x²)^½=(a²+1)/2a x+√(1+x²) = a but a = 7^⅓ so (7^⅓)³ = 7