How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$?

Question

Can anyone tell me how to solve this?

$x^6-x^5+x^4-x^3+x^2-x+1=0$

What I got to was $x^7+1=0$.

Thanks in advance.

Answer 1

$$x^6-x^5+x^4-x^3+x^2-x+1=0 \Longleftrightarrow$$ $$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0 \Longleftrightarrow$$ $$x^7+1=0 \Longleftrightarrow$$

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This introduces the extraneous root of $x=-1$, so from now on we assume that $x\ne -1$:

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$$x^7+1=0 \Longleftrightarrow$$ $$x^7=-1 \Longleftrightarrow$$ $$x^7=e^{\pi i} \Longleftrightarrow$$ $$x=\left(e^{(\pi+2\pi k) i}\right)^{\frac{1}{7}} \Longleftrightarrow$$ $$x=e^{\frac{1}{7}(\pi+2\pi k) i}$$

With $k\in\mathbb{Z}$ and $k:0-6$

Answer 2

You got the right expression. $x^7+1=0$

The roots of the equation will be like $$x= \cos (\frac{2k\pi}{7})+ i\sin (\frac{2k\pi}{7}) , 0\le k\le6$$

Note: Stress on be like, this formula will not give you the exact roots. You have to change it a bit to account for the roots of negative unity.

EDIT: Complete solution follows: $$x^7+1=0$$ or, $$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0$$ This implies your equation has exactly 1 real root and 3 pairs of complex conjugate roots.
Hence now I can write $$x^7+1=0$$ or, $$x^7=-1$$ or, $$x^7=\cos \pi + i\sin \pi = \cos (2k+1)\pi + i\sin (2k+1)\pi , 0 \le k\le 6 $$ or, $$x=[\cos (2k+1)\pi + i\sin (2k+1)\pi]^{\frac{1}{7}} , 0 \le k\le 6$$ or, $$x=\cos \frac{(2k+1)\pi}{7}+ i\sin \frac{(2k+1)\pi}{7} , 0 \le k\le 6$$

From the comment by Macavity: However,you have a spurious root included - $k=3$ . By multiplying by $x+1$ you introduced this root which is not a root of the original polynomial. So there are no real roots for the polynomial, only complex ones. Hence the final solutions are as follows: $$x=\cos \frac{(2k+1)\pi}{7}+ i\sin \frac{(2k+1)\pi}{7} , 0 \le k\le 6 \,\ \text{and} \,\ k \not = 3$$

Answer 3

This equation has the same coefficients read backwards.

There is a technique of solving such equations:

If the degree is odd, then $-1$ is a root, and dividing by $x+1$ gives you an even degree equation with the same property.

For even degree: divide by $x$ to the power half degree, and make the substitution $t=x+\frac{1}{x}$.

In this case we get $$x^3+\frac{1}{x^3}-(x^2+\frac{1}{x^2})+x+\frac{1}{x}-1=0$$

we have $$t=x+\frac{1}{x} \\ t^2=x^2+\frac{1}{x^2}+2\\ t^3=x^3+\frac{1}{x^3}+3t$$

Therefore, your equation becomes $$t^3-3t-t^2+2+t-1=0\\ t^3-t^2-2t+1=0$$

You can solve this by using the cubic formula, and then solve the corresponding quadratics.

Answer 4

We have that $x$ is a complex number, so it can be put in polar form: That is, denoting $|x|=x\bar{x}=r$, and $\ln(x/|x|)=i\phi$, we have $x=e^{i\phi}r$. Now we compute $-1=x^y=(e^{i\phi}r)^7=r^7e^{7i\phi}$, where the last step used De Moivre's formulat to derived $e^{7i\phi}=(e^{i\phi})^7$. We thus have $1=|-1|=|r^7e^{7i\phi}|=|r|^7(e^{7i\phi}e^{-7i\phi}=r^7$, so $r=1$. By Euler's formula, $e^{7i\phi}=i\sin(7\phi)+cos(7\phi)=-1$, so $sin(7\phi)=0$ and $\cos(7\phi)=-1$, so that $7\phi=n2\pi+\pi$, so $\phi=(2n+1)\pi/7$. This gives us solutions $x=e^{(2n+1)\pi/7}$.

Answer 5

As it can be deduced, the roots are complex.

A simpler approach would be to put x= -t. The roots of the corresponding equation in t will be negative of the roots of the equation in t.

On doing so we end up with t^7-1=0

Of which the roots are the seventh roots of unity.

Thus the roots of x are the negative of the seventh roots of unity

Edit: this is not nearly written as I was in a hurry(and I don't know to type equations in LaTeX). Apologies for everything.