Is it possible to simplify this expression in Mathematica?

Question

Let us consider an expression $$\frac{\sqrt[3]{\sqrt{2-x^2}+x} \sqrt[6]{1-x \sqrt{2-x^2}}}{\sqrt[3]{1-x^2}}.$$

Its plot

Plot[Surd[x + Sqrt[2 - x^2], 3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3], {x, -Sqrt[2], Sqrt[2]}, PlotRange -> All]

clearly shows this is Piecewise[{{2^(1/6),x>=-Sqrt[2]&&x<-1||x>-1&&x<1},{-2^(1/6),x>1&&x<=Sqrt[2]}}].

However, my attempts of its simplification

FullSimplify[Surd[x + Sqrt[2 - x^2], 3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3], Assumptions -> x > 1]

and

FullSimplify[Surd[x + Sqrt[2 - x^2], 3]* Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3],Assumptions ->( x >= -Sqrt[2] && x < -1)||(x>-1&&x<1)]

fail.

Knowing the result by substitution x==0 and x==Sqrt[2], the simplification can be established by

Reduce[Surd[x + Sqrt[2 - x^2], 3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3] == 2^(1/6), x, Reals]

-Sqrt[2] <= x < -1 || -1 < x < 1

and

Reduce[Surd[x + Sqrt[2 - x^2], 3]*Surd[1 - x*Sqrt[2 - x^2], 6]/ Surd[1 - x^2, 3] == -2^(1/6), x, Reals]

1 < x <= Sqrt[2]

Is there another way to simplify it in Mathematica?

Answer 1

As you have noticed, Reduce is more suitable in this case. But one does not need to know the answer in advance. Can be done as follows.

Define an expression to simplify

f = Surd[x + Sqrt[2 - x^2], 3] Surd[1 - x Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3]

Perform simplification

FullSimplify[Reduce[y == f && #, x, Reals] & /@ FunctionDomain[f, x]]

$$\left(y+\sqrt[6]{2}=0\land 1<x\leq \sqrt{2}\right)\lor \left(y=\sqrt[6]{2}\land \left(-1<x<1\lor -\sqrt{2}\leq x<-1\right)\right)$$

Answer 2

Here is my answer to the question. We start from

FunctionDomain[Surd[x+Sqrt[2-x^2],3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3],x,Reals]

-Sqrt[2] <= x < -1 || -1 < x < 1 || 1 < x <= Sqrt[2]

Therefore, the domain consists of three intervals. On every one of these intervals the function under consideration is constant. This follows from

FullSimplify[D[Surd[x + Sqrt[2 - x^2], 3]*Surd[1 - x*Sqrt[2 - x^2], 6]/Surd[1 - x^2, 3], x]]

Piecewise[{{0, x*Sqrt[2 - x^2] <= 1}}, Indeterminate]

and

Reduce[x Sqrt[2 - x^2] <= 1, x, Reals]

-Sqrt[2] <= x <= Sqrt[2]

and the mean value theorem. The values of these constants can be found by substitutions x==-Sqrt[2] and x==0 and x==Sqrt[2].

Addition. Another approach consists in

Resolve[ ForAll[x, x > 1 && x <= Sqrt[2], f == c], Reals]//ToRadicals

c == -2^(1/6)

  Resolve[ ForAll[x, x < -1 && x >= -Sqrt[2], f == c], Reals]//ToRadicals

c == 2^(1/6)