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There are different ways to embark on this problem. One way, perhaps not so "straight forward", is to set $x^6=2t^6$ for now to get $(t^3-1)(t^3+1)=0$. This of course, is extremely easy to solve, because now it is broken down to two easy cubics on which the sum\difference formulas apply. With $x=t(2^\frac{1}{6})$ you can backsub. This method avoids dealing with all sorts of higher order radicals, because of the initial substitution. You just "paste" the sixth root of $2$ as part of your final six answers. DeMoivre is avoided in case you are not familiar with that theorem
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The solution can be found as follows.
Any complex number $x$ can be written in the form
$$x=r\exp\left(i\phi+2n\pi\right)$$
where $n$ is any integer. In your equation, we have
$$x^6=2$$
or
$$r^6\exp\left(6(i\phi+2n\pi)\right)=2$$
taking the absolute value of both sides, we find
$$r^6=2$$
and thus
$$r=\sqrt[6]{2}$$
Now we require that
$$\exp\left(6(i\phi+2n\pi)\right)=\exp\left(2k\pi\right)$$
where $k$ is some other integer. Thus
$$6i\phi +12n\pi=2k\pi$$
or
$$\phi=-\frac{i}{3}\pi(k-6n)$$
but of course $k-6n$ is just some other integer $m$, so we have
$$\phi=-\frac{i}{3}\pi m$$
Now we have
$$x=\sqrt[6]{2}\exp\left(-\frac{i}{3}m\pi\right),\qquad 0\leq m \leq 6$$
Of course, any integer value for $m$ works here, but there are only $6$ unique solutions. These are your sixth roots of $2$.
On
I tend to address this and similar problems as follows:
We are given that
$x^6 - 2 = 0; \tag 1$
we want to find all such $x \in \Bbb C$; to this end we may in fact take a more general approach and find all the zeroes of
$x^n - a = 0, \; n \in \Bbb N, a \in \Bbb C; \tag 2$
we begin by writing (2) in the form
$x^n = a, \tag 3$
and expressng $a$ in polar coordinates:
$a = se^{i\phi}, \; 0 < s \in \Bbb R, \phi \in [0, 2\pi), \tag 4$
noting that we may assume $s > 0$; if $s = 0$, then $a = 0$, and it is in this case easy to see that (2) has the single root $x = 0$ of multiplicity $n$. We may also express $x$ in polars:
$x = re^{i\theta}, \tag 5$
whence
$x^n = r^ne^{in\theta}; \tag 6$
we now write (3) in terms of (4), (5) and (6):
$r^ne^{in\theta} = se^{i\phi}; \tag 7$
then
$r^n = \vert r^n \vert \vert e^{in\theta} \vert = \vert r^ne^{in\theta} \vert = \vert se^{i\phi} \vert = s \vert e^{i\phi} \vert = s, \tag 8$
so
$r = \sqrt[n] s = s^{1/n}; \tag 9$
we may divide (7) through by (8) to obtain
$e^{in\theta} = e^{i\phi}, \tag{10}$
that is,
$e^{i(n\theta - \phi)} = 1, \tag{11}$
which implies that
$i(n\theta - \phi) = 2ki\pi,\; k \in \Bbb Z \tag{12}$
and
$n\theta - \phi = 2k\pi,\; k \in \Bbb Z; \tag{13}$
we now invoke the Euclidean division algorithm, taking $n$ to be the divisor and $k$ to be the dividend, yielding
$k = nm + j, \; m \in \Bbb Z, 0 \le j < n; \tag{14}$
we substitute this into (13):
$n\theta - \phi = 2(nm + j)\pi = 2nm\pi + 2j\pi, \; m \in \Bbb Z, 0 \le j < n; \tag{15}$
$\theta - \dfrac{\phi}{n} = 2m\pi + \dfrac{2j\pi}{n}, \; m \in \Bbb Z, 0 \le j < n; \tag{16}$
$\theta = \dfrac{\phi}{n} + \dfrac{2j\pi}{n}+ 2m\pi, \; m \in \Bbb Z, 0 \le j < n; \tag{17}$
having found both $r$ (9), and the possible values of $\theta$ (17), we also have, following (5),
$x = s^{1/n}e^{i(\phi/n+ + 2\pi j/n + 2m\pi)}$ $= s^{1/n}e^{i(\phi/n+ + 2\pi j/n)}e^{2mi\pi} = s^{1/n}e^{i(\phi/n+ + 2\pi j/n)}, \; 0 \le j < n; \tag{18}$
this formula indicates that there $n$ distinct possible values of $x$, in accord with the fundamental theorem of algebra as it pertains to (2). We affirm this is so via the following
Check:
$x^n = (s^{1/n}(e^{i(\phi/n+ + 2\pi j/n)})^n = (s^{1/n})^n(e^{i(\phi/n+ + 2\pi j/n)})^n$ $= se^{i\phi}e^{2i \pi j} = se^{i\phi}(e^{2i\pi})^j = se^{i\phi}1^j = se^{i\phi} = a. \tag{19}$
We may now apply these formulas to (1), taking
$s = 2, \; \phi = 0, \; n = 6, \tag{20}$
and we finally obtain
$x = 2^{1/6} e^{2i\pi j/6} = 2^{1/6}e^{i j \pi / 3}, \; 0 \le j < 6. \tag{21}$
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With $\sqrt[3]2z=x^2$, we have a cubic equation
$$z^3-1=0$$ which we can factor as (using a remarkable product)
$$(z-1)(z^2+z+1)=0.$$
Besides the obvious root $z=1$, we also have
$$z=\frac{-1\pm i\sqrt 3}{2}.$$
Finally, the six roots are
$$x=\pm\sqrt[6]2,\pm\sqrt[6]2\sqrt{\frac{-1\pm i\sqrt 3}{2}}.$$
Note that $(1\pm i\sqrt3)^2=2(-1\pm i\sqrt3)$ and we can unnest
$$\sqrt{\frac{-1\pm i\sqrt 3}{2}}=\frac{1\pm i\sqrt3}2.$$
On
It works to substitute $z=r\exp(i\theta)$ into the equation and then solve for $r$ and $\theta$. But an easier way is to simply consider the modulus and the argument separately. Hence $$z^6=2\implies |z^6|=|2|\implies |z|^6=2\implies |z|=2^{1/6}.$$ That's half the problem done! On the other hand, $$ z^6=2\implies\arg\left(z^6\right)=\arg(2)\implies6\arg(z)=0 $$ which implies that $\arg(z)$ can be either $\pi k/3$ for any integer $0\leq k<6$. This of course means: $$ z=2^{1/6}\exp(i2\pi k/3),\quad0\leq k<6,k\in\mathbb Z.$$ With enough practice, this whole process can be done mentally and you would be able to solve such questions in one step! Unfortunately, this doesn't quite generalise easily to anything except equations of the form $z^a=b$.
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As commented above, you used $a^3-b^3=(a-b)(a^2+ab+b^2)$. You can also use $a^3+b^3=(a+b)(a^2-ab+b^2)$ and easily solve the quadratic equations: $$x^{6} - 2 = (x^{3}-\sqrt{2})(x^{3}+\sqrt{2})=\color{red}{(x-2^{1/6})}\color{green}{(x^{2}+2^{1/6}x+2^{1/3})}\color{blue}{(x+2^{1/6})}\color{purple}{(x^{2}-2^{1/6}x+2^{1/3})} \Rightarrow \\ \color{red}{x_1=2^{1/6}},\color{blue}{x_2=-2^{1/6}},\\ \color{green}{x^2+2^{1/6}x+2^{1/3}}=0 \Rightarrow x_{3,4}=\frac{-2^{1/6}\pm \sqrt{-3\cdot 2^{1/3}}}{2}=\frac{-2^{1/6}\pm 2^{1/6}\sqrt{3}i}{2},\\ \color{purple}{x^2-2^{1/6}x+2^{1/3}}=0 \Rightarrow x_{3,4}=\frac{2^{1/6}\pm \sqrt{-3\cdot 2^{1/3}}}{2}=\frac{2^{1/6}\pm 2^{1/6}\sqrt{3}i}{2}. $$
Yes. There's a result asserting that all $n$-th roots of complex number are obtained as the product of one of them by all the $n$-th roots of unity. Hence here, the sixth roots of $2$ are $$2^{\tfrac16}\mathrm e^{\tfrac{ik\pi}3},\quad 0\le k <6.$$