2016/015) Solve for x $(5+2\sqrt 6)^{x^2-3} + (5-2\sqrt 6)^{x^2-3} = 10$

We have $(5+ 2 \sqrt6)(5-2\sqrt6) = 25 - 24 =1$
So if $t= 5 + 2 \sqrt 6$ then $\dfrac{1}{t} = 5 - 2\sqrt 6$
So we get
$t^{x^2-3} + \dfrac{1}{t^{x^2-3}} = 10$
let $t^{x^3-3} = p\cdots(1) $
so we get
$ p + \dfrac{1}{p} = 10$
or $p^2 - 10p +1 = 0$
or $p = 5 + 2 \sqrt 6 $ or $p= 5 - 2\sqrt 6= (5+ 2 \sqrt 6)^{-1}$
Hence from (1) and above $x^2-3 =1  => x = \pm 2$
or $x^2 -3 = -1 => x = \pm \sqrt 2$