If you mean
x^7 + x^4 + x^3 + 1 = 0
this problem is discussed here:
www.freemathhelp.com/forum/threads/84846-SPLIT-Solve-equation-using-De-Moivre-s-theorem
Solve the equation using de moivre's theorem (1) xto the power 7+x to the power 4+x to the power 3+1=0
2 answers
Did you notice that the equation factors nicely using grouping ?
x^4(x^3 + 1) + x^3 + 1 = 0
(x^4 + 1)(x^3 + 1) = 0
x^3 + 1 is the sum of cubes, and you should know the factors for it.
x^3 + 1 = (x+1)(x^2 - x + 1) = 0
x = -1 or x = (1 ± √-3)/2 = (1 ± √3 i)/2
so there is no need to use De Moivre's on this one, but I will leave it up to you to apply it and get the same answers.
let's for the x^4+1 = 0
let x^4 = -1 = -1 + 0i , = 1(cos180° + i sin180°)
x = (1(cos180° + i sin180°))^(1/4)
= 1(cos 45° + i sin 45° = √2/2 + √2/2 i <------ the primary fourth root
but 360/4 = 90, so adding 90° will yield the other 3 roots
x = cos(45+90) + isin(45+90) = cos135 + isin135 = -√2/2 + √2/2 i
x = cos225 + isin225 = -√2/2 - √2/2 i
x = cos315 + isin315 = √2/2 - √2/2 i
so x = -1, (1 ± √3 i)/2 , ±√2/2 ± √2/2 i <-----7 roots
x^4(x^3 + 1) + x^3 + 1 = 0
(x^4 + 1)(x^3 + 1) = 0
x^3 + 1 is the sum of cubes, and you should know the factors for it.
x^3 + 1 = (x+1)(x^2 - x + 1) = 0
x = -1 or x = (1 ± √-3)/2 = (1 ± √3 i)/2
so there is no need to use De Moivre's on this one, but I will leave it up to you to apply it and get the same answers.
let's for the x^4+1 = 0
let x^4 = -1 = -1 + 0i , = 1(cos180° + i sin180°)
x = (1(cos180° + i sin180°))^(1/4)
= 1(cos 45° + i sin 45° = √2/2 + √2/2 i <------ the primary fourth root
but 360/4 = 90, so adding 90° will yield the other 3 roots
x = cos(45+90) + isin(45+90) = cos135 + isin135 = -√2/2 + √2/2 i
x = cos225 + isin225 = -√2/2 - √2/2 i
x = cos315 + isin315 = √2/2 - √2/2 i
so x = -1, (1 ± √3 i)/2 , ±√2/2 ± √2/2 i <-----7 roots