I met a algebraic equation(not a transcendent equation) during my study of Stoner criterion in Quantum Statistical Physics. In this occasion, one need to solve the equation $$ (\xi+1)^{2/3}-(\xi-1)^{2/3}=\frac43a\xi $$ where $a$ is a parameter. Here below we assume that $a>1$ of which the equation has two non-zero roots $x$ and $x'$. Since $x+x'=0$, we just focus on the positive one.
However, possibly due to the lack of some mathematical techniques, it is hard for us to solve the equation analytical even up to now. Some pepople suggested an approximation formula $$ \xi=3.79(a-1)^{0.504}-5.72(a-1)^{1.304} $$ which is rather precision when $a$ is not too big.
My question is that can we express the root by some known functions? Such as Lambert $W$ function. However, any approximation formula will also be welcomed. You can tell me you way to treat this problem, or some references instead. Thank you very much.
It's not transcendental, it's algebraic. In particular, $\xi$ satisfies the polynomial
$$ 4096 a^9 \xi^6-20736 a^6 \xi^4+19683 a^3 \xi^4-4374 a^3 \xi^2+19683 a^3-19683 = 0 $$
Note that when $a = a_0 = 3\times 2^{-4/3} \approx 1.190550789$, this factors as $$ \dfrac{19683}{16} (16 \xi^2 + 11)(\xi - 1)^2 (\xi + 1)^2$$
The two roots near $1$ can then be expanded in a Puiseux series around $a=a_0$ (real for $a < a_0$) as follows:
$$\xi = 1 \pm \dfrac{8}{9} \sqrt{3} (a_0 - a)^{3/2} - \dfrac{16}{9} 2^{2/3} (a_0 - a)^{2} \pm \dfrac{224}{81} 2^{1/3} 3^{1/2} (a_0 - a)^{5/2} -\dfrac{7648}{729} (a_0 - a)^3 + \ldots $$
The root with $+$ solves your original equation there.