Polynomial roots problems. $$ X^5-55X+21$$
Prove that the given polynomial has 2 roots which satisfy the condition:
$$X_1X_2=1$$ 
and find them.
I have tried to make use of Viette's relations ,but couldnt get to a satisfying result.
 A: The solution needn't be pulled out of hat like magic. Rather, we can discover it by exploiting innate reciprocation symmetry, i.e. if both $\,x_1$ and $\,x_2 = x_1^{-1}$ are roots of $f(x)$ then
$ 0 = f(x_2) = f(x_1^{-1})\,\Rightarrow\,0 = x_1^5f(x_1^{-1}) = 1\!-55x_1^4\!+21x_1^5 = \tilde f(x_1) = \,$ reverse of $f(x_1)$
Therefore $\,x_1\,$ is a root of $\,g(x) := \gcd(f(x),\tilde f(x))\,$ since $\,g(x)  = a(x) f(x) + b(x) \tilde f(x)$
by the Bezout gcd identity. By Euclid the gcd $= x^2\!-3x +\color{#c00}1$ whose roots have product $= \color{#c00}1$.
Remark $\ $ Such inversion symmetry often proves useful, e.g. see here.
Generally it is wise to attempt to  exploit any innate symmetry before diving head-first into difficult brute force calculations. For example, for problems of the above sort it is much more efficient to compute polynomial gcds than to compute polynomial factorizations. Follow the prior link for many more examples of such symmetry exploits.
A: $$ x^5-55x+21=(x^2-3x+1)(x^3+3x^2+8x+21)$$
It's not hard to prove that $x^3+3x^2+8x+21$ has only 1 real root, negative (just for curiosity, we won't need him), while $x^2-3x+1$ has 2 positive roots: $\frac{3\pm\sqrt{5}}{2}$.
It turns out that these 2 roots are good for us:
$$\frac{3+\sqrt{5}}{2}\cdot\frac{3-\sqrt{5}}{2}=1$$
A: Hint: Check the polynomial GCD of the polynomial and its reverse, $21X^5-55X^4+1$.
