Solve the equation $\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3}{x+\sqrt2}}$ Solve the equation $$\dfrac{\sqrt[7]{x-\sqrt2}}{2}-\dfrac{\sqrt[7]{x-\sqrt2}}{x^2}=\dfrac{x}{2}\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$ We have $x\ne0;-\sqrt2$.
Let's multiply both sides of the equation by $2x^2\ne0$ to get $$x^2\sqrt[7]{x-\sqrt2}-2\sqrt[7]{x-\sqrt2}=x^3\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$$$(x^2-2)\sqrt[7]{x-\sqrt2}=x^3\sqrt[7]{\dfrac{x^3}{x+\sqrt2}}$$ Let's multiply both sides of the equation by$\sqrt[7]{x+\sqrt2}\ne0$ to get $$(x^2-2)\sqrt[7]{x^2-2}=x^3\sqrt[7]{x^3}$$ $$(x^2-2)^8=x^{24}$$
 A: Hint
Write the equation as
$$\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3(x-\sqrt2)}{x^2-2}}$$
In the transformation above, take care of the situation $x=\sqrt 2$.
$$\frac{\sqrt[7]{x-\sqrt2}}{2}-\frac{\sqrt[7]{x-\sqrt2}}{x^2}=\frac{x}{2}\sqrt[7]{\frac{x^3(x-\sqrt2)}{x^2-2}}$$
$$\left(\sqrt[7]{x-\sqrt2}\right)\left(\frac{1}{2}-\frac{1}{x^2}-\frac{x}{2}\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\frac{1}{2x^2}\left(\sqrt[7]{x-\sqrt2}\right)\left(x^2-2-x^3\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\frac{1}{2x^2}\left(\sqrt[7]{x-\sqrt2}\right)\left(x^2-2-x^3\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\frac{x^2-2}{2x^2}\left(\sqrt[7]{x-\sqrt2}\right)\left(1-\frac{x^3}{x^2-2}\sqrt[7]{\frac{x^3}{x^2-2}}\right)=0$$
$$\left(\frac{x^2-2}{2x^2}\right)\left(\sqrt[7]{x-\sqrt2}\right)\left(1-\left(\frac{x^3}{x^2-2}\right)^{8/7}\right)=0$$
A: Starting where you left off, at $(x^2-2)^8=x^{24}$, rearrange the equation to:
$$x^{24} - (x^2-2)^8 = 0$$
You could use the Binomial Theorem to expand the $(x^2-2)^8$ term, but for now I won't.  Just note that the polynomial's constant term (and the product of all its roots) is $-(-2)^8 = -256$.
If you just want real solutions
By the Rational Root Theorem, any rational roots of the polynomial must be factors of 256, i.e., $x \in \pm\{ 1, 2, 4, 8, 16, 32, 64, 128, 256 \}$.  It turns out that $x = 1$ and $x = -1$ are the only ones that work, and then there are no other real solutions.
If you want complex solutions
The exponents are even, so the 24th-degree polynomial can be factored as a difference of two squares:
$$(x^{12} + (x^2-2)^4)(x^{12} - (x^2-2)^4) = 0$$
And we can do this two more times:
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2)(x^6 - (x^2-2)^2) = 0$$
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2)(x^3 + (x^2-2))(x^3 - (x^2-2)) = 0$$
Or, un-nesting the parentheses in the cubics:
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2)(x^3 + x^2 - 2)(x^3 - x^2 + 2) = 0$$
You can use the Rational Root Theorem to find one factor of each of the two cubics, and then the quadratic formula to find the other two:

*

*$x^3 + x^2 - 2 = 0 \implies (x-1)(x^2 + 2x + 2) = 0 \implies x \in \{ 1, -1 + i, -1 - i \}$

*$x^3 - x^2 + 2 = 0 \implies (x+1)(x^2 - 2x + 2) = 0 \implies x \in \{ -1, 1+i, 1-i  \}$
This gives us 6 of the original polynomial's 24 roots.  To find others, we can use $a^2+b^2 = a^2-i^2b^2 = (a+ib)(a-ib)$.
$$(x^{12} + (x^2-2)^4)(x^6 + (x^2-2)^2) = 0$$
$$(x^{6} + i(x^2-2)^2)(x^{6} - i(x^2-2)^2)(x^3 + i(x^2-2))(x^3 - i(x^2-2)) = 0$$
This gives us two more cubics.  There is a cubic formula that can find the roots, but they'll be messy nested-radical things.
Finally, for the sextic polynomials, use $\sqrt{i} = \pm \frac{1 + i}{\sqrt 2}$ to do the sum/difference of squares thing one more time.  Again, I'm not going to even try to write the radical forms.
