We first notice that the given equation in a cubic equation in terms of $\sqrt[3]{x}$, $\sqrt[3]{2x+6}$ and $\sqrt[3]{3x+24}$. Let $a = \sqrt[3]{x}$, $b = \sqrt[3]{2x+6}$, and $c = \sqrt[3]{3x+24}$. Thus, the equation can be written as:

$a + b = c$

Cubing both sides, we get:

$a^3 + b^3 + c^3 + 3(ab(a+b+c)) = a^3 + b^3 + c^3 + 3abc$

Since $a = \sqrt[3]{x}$, $b = \sqrt[3]{2x+6}$, and $c = \sqrt[3]{3x+24}$, we substitute these values in:

$x + 2x + 6 + 3x + 24 + 3\left(\sqrt[3]{x}\right)\left(\sqrt[3]{2x+6}\right)\left(\sqrt[3]{3x+24}\right) = x + 2x + 6 + 3\sqrt[3]{x}\sqrt[3]{2x+6}\sqrt[3]{3x+24}$

Simplifying, we get:

$x + 2x + 6 + 3x + 24 + 3\sqrt[3]{x(2x+6)(3x+24)} = x + 2x + 6 + 3\sqrt[3]{x(2x+6)(3x+24)}$

Combining like terms, we get:

$6x + 30 + 3\sqrt[3]{6x^3 + 72x^2 + 144x} = 6 + 3\sqrt[3]{6x^3 + 72x^2 + 144x}$

Subtracting $6$ from both sides, we have:

$6x + 30 + 3\sqrt[3]{6x^3 + 72x^2 + 144x} - 6 = 3\sqrt[3]{6x^3 + 72x^2 + 144x}$

$6x + 24 + 3\sqrt[3]{6x^3 + 72x^2 + 144x} = 3\sqrt[3]{6x^3 + 72x^2 + 144x}$

Subtracting $3\sqrt[3]{6x^3 + 72x^2 + 144x}$ from both sides, we get:

$6x + 24 = 0$

This implies $x=-4$. Therefore, the solution to the equation $\sqrt[3]{x} +\sqrt[3]{2x+6} =\sqrt[3]{3x+24}$ is $\boxed{x=-4}$.