How to find the complex solution of $x^6$

Question

How do you find the complex solutions to $x^6+x^3-2=0 $

Obviously $x=1$ is one solution, but i cant get further than that.

Answer 1

Substitute $z=x^3$. Then you can find $z$ easily,and eventually $x$.

Answer 2

$$x^6+x^3-2=0$$ $$(x^3-1)(x^3+2)=0$$ We'll now seperately solve these factors $$z^3=1$$ $$z^3=\cos(2k\pi)+i\sin(2k\pi)$$ $$z=\left[\cos(2k\pi)+i\sin(2k\pi)\right]^{1/3}$$ Using De-Movires Theorem $$z=\cos\left(\frac{2k\pi}{3}\right)+i\sin\left(\frac{2k\pi}{3}\right)$$

$$z=1,e^{2i\pi/3},e^{4i\pi /3}.$$

Similarly $$z^3=2\cdot\left[\cos(2k\pi)+i\sin(2k\pi)\right]$$ $$z=\sqrt[3]{2}\cdot\left[\cos((2k+1)\pi)+i\sin((2k+1)\pi)\right]^{1/3}$$ Using De-Movires Theorem $$z=\sqrt[3]{2}\cdot\cos\left(\frac{(2k+1)\pi}{3}\right)+i\sin\left(\frac{(2k+1)\pi}{3}\right)$$

$$z=\sqrt[3]{2}\cdot e^{ i\pi/3}, \sqrt[3]{2}\cdot e^{i\pi }, \sqrt[3]{2}\cdot e^{5i\pi /3}.$$

All six roots of $x^6+x^3-2=0$ are

$$z=1 , e^{2i\pi/3} , e^{4i\pi /3} , -\sqrt[3]{2} , \sqrt[3]{2}\cdot e^{ i\pi/3} , \sqrt[3]{2}\cdot e^{5i\pi /3}$$

Answer 3

$$z:=x^3 \iff z^2+z-2=0 \iff (z-1)(z+2)=0 \iff x^3-1=0 \;\; \text{or} \;\; x^3+2=0$$

$$x^3-1=0 \iff (x-1)(x^2+x+1)=0$$

$$x^3+2=0 \iff (x+\sqrt[3]{2})(x^2-\sqrt[3]{2}x+\sqrt[3]{4})=0$$

Answer 4

So our equations is:

$$x^6 + x^3 -2 =0$$

If we make a substitution $u = x^3$

$$u ^2 + u - 2 = 0$$

Now we have a simple quadratic we can factor.

$$(u+2)(u-1) = 0$$

$$u = -2, 1$$

So we have 2 cases.

$$x^3 = -2$$

Representing $-2$ in polar form:

$$r = \sqrt{2^2} = 2$$

$$\cos(\theta) = -1,\ \theta = \pi$$

So our polar form is:

$$-2 = 2\left(\cos(\pi) + i\sin(\pi)\right)$$

So,

$$(-2)^{1/3} = \left(2\left(\cos(\pi) + i\sin(\pi)\right)\right)^{1/3} = 2^{1/3}\left(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})\right)$$

Another solution is:

$$2^{1/3}\left(\cos(\frac{\pi}{3}) - i\sin(\frac{\pi}{3})\right)$$

The other case is when $x^3 = 1$

So then we have $x^3 -1 = 0, (x-1)(x^2+x+1) = 0$

Therefore one solution is $x = 1$

The others are:

$$x = \frac{-1 + \sqrt{1 - 4}}{2} = \frac{-1 + i\sqrt{3}}{2},\ \frac{-1 - i\sqrt{3}}{2} $$

Therefore all our $6$ solutions are found.

Answer 5

$$(x^3)^2 + x^3 - 2= 0\\ \iff (x^3 - 1)(x^3 + 2) = 0\\ \iff (x - 1)(x^2 + x + 1)(x+2^{1/3})(x^2 - x + 2^{2/3}) = 0 $$

Use the quadratic equations for the quadratics and solve.

The solutions are hence: $$x = 1, -2^{1/3} , \frac{-1 \pm i\sqrt{3}}{2}, \frac{1 \pm i\sqrt{2^{2/3}-1}}{2}$$

Out of our $6$ solutions, $2$ are real while $4$ are complex.