An algebra problem by Albin Jonsson

$\begin{aligned} (\color{#3D99F6}{7x-2})^{\frac{1}{3}} + (7x+5)^{\frac{1}{3}} & = 3 \quad \quad \small \color{#3D99F6}{\text {Let }y = 7x-2} \\ \implies \color{#3D99F6}{y}^{\frac{1}{3}} + (\color{#3D99F6}{y}+7)^{\frac{1}{3}} & = 3 \\ (y+7)^{\frac{1}{3}} & = 3 - y^\frac{1}{3} \quad \quad \small \color{#3D99F6}{\text{Cubing both sides}} \\ y + 7 & = 27 - 27y^\frac{1}{3} + 9y^\frac{2}{3} - y \\ 2y - 9y^\frac{2}{3} + 27y^\frac{1}{3} - 20 & = 0 \quad \quad \small \color{#3D99F6}{\text{By rational root theorem}} \\ \left(y^\frac{1}{3} - 1\right) \left(2 y^\frac{2}{3} -7y^\frac{1}{3} +20 \right) & = 0 \\ \implies y^\frac{1}{3} & = 1 \quad \quad \small \color{#3D99F6}{\text{The only real root}} \\ y & = 1 \\ 7x - 2 & = 1 \\ \implies x & = \frac{3}{7} \approx \boxed{0.428} \end{aligned}$

The only method that I can think of to solve this is Newton Raphson Method. Anyone got a better solution?

Anyway, let $f(x) = (7x-2)^{1/3} + (7x+5)^{1/3} - 3$

We know that the graph of cube root functions are continuous, and given that:

$f(1) = 5^{1/3} + 12^{1/3} - 3 \approx 0.999\\ f(0) = (-2)^{1/3} + 5^{1/3} - 3 \approx -2.550$

From here, we can use the intermediate value theorem to deduce that a root exists in the interval $(0,1)$

Now, use the formula below to find the root:

$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)}$

Let $x_0 = 1$

$f'(x) = \dfrac{7}{3} \left( (7x-2)^{-2/3} + (7x+5)^{-2/3} \right)$

$x_1 = 1 - \dfrac{f(1)}{f'(1)} \approx 0.19607 \quad\quad$ (I'm lazy to type out the whole thing, you do it yourself)

$x_2 = x_1 - \dfrac{f(x_1)}{f'(x_1)} \approx 0.71444$

The remaining iterations are:

$x_3 \approx 0.34690\\ x_4 \approx 0.40942\\ x_5 \approx 0.42781\\ x_6 \approx 0.42857\\ x_7 \approx 0.42857$

Therefore, the value of $x$ that satisfies this equation is $x=\boxed{0.42857}$

Notes: The question never stated how many iterations or how many decimal places should our answer be. If it's 3 decimal places, wouldn't it be $0.429$ instead of $0.428$ ?

And if there is no other method to solve this, does this question belong to Calculus instead?