I've been really struggling to solve this one, could you provide how you'd solve it?
$$3x^{2/3} + 4x^{1/3} =4$$
Solve this equation with rational exponents
radical-equations
Related Solutions
The answer is no, on the assumption that any process of "simplification" will yield a polynomial in $x$ with rational coefficients if the coefficients of the given quadratics are rational.
Consider any case where $h(x) = m(x)$ and $f(x) = 0$.
If $x$ satisfies the given equation, then $$ \sqrt{g(x)} - \sqrt{l(x)} = k(x) $$ and therefore $$ \tag{$*$}\label{eq:1} 4g(x)l(x) = (k(x)^2 - g(x) - l(x))^2. $$ Now specialise further, to the case: \begin{align*} g(x) & = x^2 + \tfrac12, \\ l(x) & = x^2 - \tfrac12, \\ k(x) & = x^2 + 1. \end{align*} \eqref{eq:1} becomes $$ 4x^4 - 1 = (x^4 + 1)^2. $$ This can be solved explicitly in terms of nested square roots, but it is enough for our purposes to rewrite it as $$ x^8 - 2x^4 + 2 = 0, $$ and observe that by Eisenstein's criterion the polynomial on the left hand side is irreducible over the rationals, therefore $x$ does not satisfy any polynomial equation of degree less than $8$ with rational coefficients.
Update
This uses a second example to answer the question of whether reduction to a quartic equation is possible. It takes into account the clarification of the question in comments made since this answer was first posted. A further update below uses a third example to answer the more general question asked in the title of the thread: does the equation have an algebraic solution (which might not be obtained by reduction to a quartic)?
Consider the special case $$ \frac{x^2 + 1 + \sqrt{x^2 + x + \tfrac12}}{x^2 + 1} = \frac{2x^2 + \sqrt{x^2 - x + \tfrac12}}{x^2 + 1}. $$ Here we have \begin{align*} g(x) & = x^2 + x + \tfrac12 = \left(x + \tfrac12\right)^2 + \tfrac14 > 0, \\ l(x) & = x^2 - x + \tfrac12 = \left(x - \tfrac12\right)^2 + \tfrac14 > 0, \end{align*} for all real $x$, so the stated conditions are satisfied. The equation simplifies to \begin{equation} \tag{$**$}\label{3380555:eq:2} \sqrt{g(x)} - \sqrt{l(x)} = n(x), \end{equation} where $n(x) = x^2 - 1$. The four equations $$ \pm\sqrt{g(x)} \pm\sqrt{l(x)} = n(x) $$ together are equivalent to a single polynomial equation $$ 4g(x)l(x) = (n(x)^2 - g(x) - l(x))^2. $$ Because $g(x)l(x) = \left(x^2 + \tfrac12\right)^2 - x^2 = x^4 + \tfrac14$, and $g(x) + l(x) = 2x^2 + 1$, the polynomial equation reduces to $$ 4x^4 + 1 = (x^4 - 4x^2)^2, $$ i.e. $p(x) = 0$, where $$ p(x) = x^8 - 8x^6 + 12x^4 - 1. $$ Write this as $p(x) = q(x^2)$, where $q(y) = y^4 - 8y^3 + 12y^2 - 1 = y^2(y - 2)(y - 6) - 1$. $$ \begin{array}{r|rrrrrr} y & -1 & 0 & 1 & 2 & 6 & 7 \\ q(y) & 20 & -1 & 4 & -1 & -1 & 244 \end{array} $$ $p(x)$ therefore has six real roots and two imaginary roots. Corresponding to the root of $q(y)$ just below $2$, $p(x)$ has the real root $$ x \bumpeq 1.390776704210124 $$ (as calculated in Python using Newton's method).
Numerical calculation confirms that this value of $x$ is also a solution of \eqref{3380555:eq:2}. (A comment below shows that this is the only positive solution.)
I didn't have the heart to try to solve the quartic equation $q(y) = 0$ to obtain a closed-form expression for $x$ and try to use it to prove that \eqref{3380555:eq:2} is satisfied! Instead one can prove it by using inequalities. We have $1 < x^2 < 2$, therefore $0 < n(x) < 1$. Also $g(x) > l(x)$, so we cannot have $-\sqrt{g(x)} \pm\sqrt{l(x)} = n(x)$. Next, $\sqrt{g(x)} > x + \tfrac12$ and $\sqrt{l(x)} > x - \tfrac12$, so $\sqrt{g(x)} + \sqrt{l(x)} > 2x > 2 > n(x)$. Now only \eqref{3380555:eq:2} remains as a possibility.
We have $p(X) \equiv X^8 + X^6 + 2 \pmod3$, and the latter polynomial is listed as being irreducible on this web site: Irreducible trinomials over GF(3). I take this to be good evidence that $p(x)$ is irreducible, although I haven't verified the result stated on the web site. It follows as before that $x$ does not satisfy any polynomial equation of degree less than $8$ with rational coefficients.
Update 2
Consider the special case $$ \frac{x^2 + 2 + \sqrt{x^2 + x + 1}}{x^2 + 2} = \frac{x^2 + 1 + \sqrt{x^2 - x + 1}}{x^2 + 1}. $$ This is equivalent to $$ (x^2 + 1)\sqrt{x^2 + x + 1} = (x^2 + 2)\sqrt{x^2 - x + 1}. $$ Because both sides of the latter equation are positive, it is in turn equivalent, for real $x$, to the polynomial equation $$ (x^2 + 1)^2(x^2 + x + 1) = (x^2 + 2)^2(x^2 - x + 1). $$ Simplifying: $$ x[(x^2 + 1)^2 + (x^2 + 2)^2] = (x^2 + 1)[(x^2 + 2)^2 - (x^2 + 1)^2]. $$ This equation simplifies in turn to the quintic equation $p(x) = 0$, where $$ p(x) = 2x^5 - 2x^4 + 6x^3 - 5x^2 + 5x - 3. $$ $p'''(x) = 120x^2 - 48x + 36 = 120\left(x-\tfrac15\right)^2 + 31\tfrac15 > 0$ for all $x$, so the cubic $p''(x)$ is strictly increasing. Its unique zero is $x_0 \bumpeq 0.30861$. But $p'(x_0) \bumpeq 3.4838$, therefore $p'(x) > 0$ for all $x$, therefore $p(x)$ is strictly increasing, therefore it has a unique real zero: $$ x \bumpeq 0.705526664475018. $$ We have $3p(X) \equiv X^5 + 4X^4 + 3X^3 + 1 \pmod{5}$, which is listed as irreducible here, implying that $p(X)$ is irreducible in $\mathbb{Q}[X]$.
According to GAP 4.8.10:
gap> GaloisType(UnivariatePolynomial(Rationals,[-3,5,-5,6,-2,2]));
5
gap> TransitiveGroup(5,5);
S5
I think this proves that the Galois group of $p(X)$ is isomorphic to $S_5$. This group is famously not solvable. Therefore, no root of $p(X)$, including its unique real root shown above, is given by an expression involving only integer constants, field operations, and/or extractions of roots.
It follows that the general equation $$ \frac{f(x) + \sqrt{ g(x)}}{h(x)} = \frac{k(x) + \sqrt{ l(x)}}{m(x)} $$ has no general solution expressed in terms of field operations, extractions of roots, and the coefficients of $f, g, h, k, l, m$.
The problem is that $(x^2)^{1/2} = \lvert x\rvert$, not $x$. So $\lvert x \rvert = 3$ and therefore $x = \pm 3$.
Best Answer
Here, notice how one term has twice the exponent of the other:
$$3x^{\color{blue}{\frac{2}{3}}}+4x^{\color{green}{\frac{1}{3}}} = 4$$
$$x^{\frac{2}{3}} = x^{2\cdot\frac{1}{3}} = \left(x^{\frac{1}{3}}\right)^2$$
So, you can rewrite the equation as
$$3\left(x^{\frac{1}{3}}\right)^2+4x^{\frac{1}{3}} = 4$$
The good thing here is that you can now make the substitution $t = x^\frac{1}{3}$ in order to reach a simple quadratic equation:
$$3t^2+4t = 4 \iff 3t^2+4t-4 = 0$$
From here, you can solve the quadratic for $t$ and finally obtain $x$. This technique is typically used for all equations of the form $ax^{2n}+bx^n+c = 0$: letting $t = x^n$ and solving $at^2+bt+c = 0$.