Is it the same $a^\alpha + b^\alpha = x$ as $(a^\alpha)^\frac{1}{\alpha} + (b^\alpha)^\frac{1}{\alpha} = x^\frac{1}{\alpha}$

Question

I want to ask if it is possible to transform equation $a^\alpha + b^\alpha = x$ into $a + b = x^\frac{1}{\alpha}$ by elevating each parameter by $\frac{1}{\alpha}$? Or if we elevate each parameter by $\frac{1}{\alpha}$ it must be $(a^\alpha + b^\alpha)^\frac{1}{\alpha} = x^\frac{1}{\alpha}$?

Answer 1

No it isn't in general since

$$x=a^\alpha + b^\alpha \implies x^\frac{1}{\alpha}=(a^\alpha + b^\alpha)^\frac{1}{\alpha}$$

For general $a$ and $b$ it is trivially true only for $\alpha=1$.