Solve the equation \(x^7+x^4+x^3+1=0\).

Answer: \(x=e^{\pm i\pi/4},\ e^{\pm i3\pi/4}\) (i.e. \(\pm\frac1{\sqrt2}\pm\frac{i}{\sqrt2}\)), \(x=-1\), \(x=\frac12\pm\frac{\sqrt3}{2}i\).

Factorise: \(x^7+x^4+x^3+1=x^4(x^3+1)+(x^3+1)=(x^4+1)(x^3+1)=0\).

Case 1: \(x^4=-1=\cos\pi+i\sin\pi=\cos(2k\pi+\pi)+i\sin(2k\pi+\pi)\).

By De Moivre's theorem, \[x=\cos\frac{(2k+1)\pi}{4}+i\sin\frac{(2k+1)\pi}{4},\quad k=0,1,2,3.\]

\(k=0:\ \frac{1}{\sqrt2}+\frac{i}{\sqrt2}\); \(k=1:\ -\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\); \(k=2:\ -\frac{1}{\sqrt2}-\frac{i}{\sqrt2}\); \(k=3:\ \frac{1}{\sqrt2}-\frac{i}{\sqrt2}\).

Case 2: \(x^3=-1=\cos(2k\pi+\pi)+i\sin(2k\pi+\pi)\).

\[x=\cos\frac{(2k+1)\pi}{3}+i\sin\frac{(2k+1)\pi}{3},\quad k=0,1,2.\]

\(k=0:\ \frac12+\frac{\sqrt3}{2}i\); \(k=1:\ \cos\pi+i\sin\pi=-1\); \(k=2:\ \frac12-\frac{\sqrt3}{2}i\).

So the seven roots are the four values from Case 1 and the three values from Case 2. (Check: the degree is 7 and we have 7 distinct roots.)