I'm looking for the roots of the sextic equation in $x$ $x^6 - (3 m) x^5 + 5 m^2 x^4 - (5 m^3) x^3 + 3 m^4 x^2 - m^5 x + L = 0$. I know that at most two of the roots of this are real when $m$ and $L$ are positive integers. Also mathematica finds a closed form for all the roots (surprisingly). They are $x = 1/2 (m - sqrt(2 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) + (4 2^(1/3) m^4)/(3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) - 5 m^2)/sqrt(3))$, $x = 1/2 (sqrt(2 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) + (4 2^(1/3) m^4)/(3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) - 5 m^2)/sqrt(3) + m)$, $x = 1/2 (m - sqrt(m^2 - 4 (((1 - i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 + i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))$, $x = 1/2 (m + sqrt(m^2 - 4 (((1 - i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 + i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))$, $x = 1/2 (m - sqrt(m^2 - 4 (((1 + i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 - i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))$, and $x = 1/2 (m + sqrt(m^2 - 4 (((1 + i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 - i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))$. What I need to know is, which two of these roots are the ones that are candidates to be real, when $m$ and $L$ are positive integers? Also, how does one determine this (without substituting values)?

Pardon the messy values, it's more worth it to just feed it into mathematica.