<s>[INST] Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
 [/INST] if $x = \sqrt{8-\sqrt{63}},$  then $0<x<1$ and $x^2 - 8 = - \sqrt{63},$ then $x^4 - 16 x^2 + 64= 63,$  then
$$ x^4 - 16 x^2 + 1 = 0.  $$ Also
$$  x^2 - 16 + \frac{1}{x^2} = 0 $$
Taking $$ u = x + \frac{1}{x} $$
we get $u^2 - 18 = 0 $ and $$  u = \sqrt {18} $$
and
$$  x = \frac{3 \sqrt 2 \pm \sqrt{14}}{2} $$
and $x<1$ gives
$$ \color{blue}{ x = \frac{3 \sqrt 2 - \sqrt{14}}{2} } $$
Let's see, I could have chosen $$  v = \frac{1}{x} - x > 0 $$
with $v^2-14 = 0,$ then $v = \sqrt{14}$ or
$$  x^2 + \sqrt{14} x - 1 = 0, $$
$$  x = \frac{- \sqrt{14} \pm \sqrt{18}}{2} $$
and $x>0$ gives
$$ \color{red}{ x = \frac{- \sqrt{14} + \sqrt{18}}{2} } $$
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