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 $\boxed{x=-4}$.
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 $\boxed{x=-4}$.