$y = (x^3 + \frac{1}{x^3}) - 6(x^2 + \frac{1}{x^2}) + 3(x + \frac{1}{x})$ find min value
$t = x + \frac{1}{x}$
$y = t^3 -6t^2 + 12$
critical point when $t = 4, x = 2 \pm \sqrt3 $
it's hard to evaluate the value of $f(2 - \sqrt3) $ manually
is there a way to evaluate it in easy way? (not allowed using calculator)
The minimal value does not exist. Try $x\rightarrow0^-.$
For $x>0$ we can use AM-GM.
Indeed, let $x+\frac{1}{x}=2u$.
Thus, $u\geq1$ and $$x^3+\frac{1}{x^3}-6\left(x^2+\frac{1}{x^2}\right)+3\left(x+\frac{1}{x}\right)=$$ $$=8u^3-6u-6(4u^2-2)+6u=8u^3-24u^2+12=$$ $$=4\left(2u^3-6u^2+3\right)=4\left(2u^3+8-6u^2-5\right)\geq4\left(3\sqrt[3]{(u^3)^2\cdot8}-6u^2-5\right)=-20.$$ The equality occurs for $u=2$, which says that we got a minimal value.