Where is the mistake in the equality of roots?

Question

I have this statement:

If $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}}, g(x) = > \frac{\sqrt[6]{8x^2}}{2}$, with $x < 0$ is $f(x) = g(x)$ ?

My attempt was:

$(1)$ $f(x) = \frac{\sqrt[3]{x}}{\sqrt{2}} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$

$(2)$ $g(x) = \frac{\sqrt[6]{8x^2}}{2}$ using the property of $\sqrt[n]{k^m} = k^{\frac{m}{n}}$ and since $x^2 \geq 0$, thus:

$(3)$ $g(x) = \frac{\sqrt[6]{2^3}\sqrt[6]{x^2}}{2} = \frac{2^{\frac{3}{6}}\cdot (x^2)^{\frac{1}{6}}}{2} = \frac{\sqrt{2}\sqrt[3]{x}}{2}$, that is equal to $f(x)$ from the step $(1)$

But according to the guide, this is false. So, what is wrong?

Answer 1

with $x<0$, $\sqrt[3]x<0$, but $\sqrt[6]{x^2}>0$

Answer 2

If $x \geq 0$, then

$$ \frac{\sqrt[6]{8x^{2}}}{2} = \frac{\sqrt[6]{2^{3}x^{2}}}{2} = \frac{2^{1/2}x^{1/3}}{2} = \frac{\sqrt[3]x}{\sqrt{2}}. $$

If $x < 0$,

$$ \frac{\sqrt[6]{8x^{2}}}{2} = \frac{\sqrt[6]{2^{3}x^{2}}}{2} = \frac{2^{1/2}|{x}\mid^{1/3}}{2} = \frac{\sqrt[3]|x|}{\sqrt{2}}. $$