Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$ I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$
$$2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7 \implies 2^{y^2}=(3^{\frac{x}{2}}-7)^2$$
$$2^{y^2}=(3^{\frac{x}{2}}-7)^2 \implies 2^{y^2}=3^{x}-14\cdot3^{\frac{x}{2}}+49$$
So putting that in the first equation:
$$3^x-2^{y^2}=77 \implies 3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77$$
$$\mathbf{3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77 \implies 14\cdot3^{\frac{x}{2}}=126}$$
$$14\cdot3^{\frac{x}{2}}=126 \implies 3^{\frac{x}{2}}=9$$
$$3^{\frac{x}{2}}=9 \implies x = 4$$
And putting the value of $x$ in the second equation we get $y$:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 3^{2}-2^{\frac{y^2}{2}}=7$$
$$3^{2}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=2$$
$$2^{\frac{y^2}{2}}=2 \implies y^2 = 2$$
$$y^2 = 2 \implies y = \sqrt{2}$$
So I tried solving the system in the question title with the same strategy until the step in bold, which gave me $3^x - 3^{\frac{2x}{3}}+14\cdot3^{\frac{x}{3}}=126$, and I have no idea how to progress further.

Using the hyperbolic method when there is only one real root as for$$u^3-u^2+14u-126=0$$ the solution is not ugly at all. It write
$$u=\frac{1}{3} \left(1+2 \sqrt{41} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{1639}{41 \sqrt{41}}\right)\right)\right)$$ while using Cardano method, it write
$$\frac{1}{3} \left(1-\frac{41}{\sqrt[3]{1639+3 \sqrt{306138}}}+\sqrt[3]{1639+3
   \sqrt{306138}}\right)$$
Which one do you prefer ?
