﻿ Given x^5+x+1=0, find the value of x^3-x^2

#### Question

Given x^5+x+1=0, find the value of x^3-x^2

Collected in the board: Quintic Equation

Steven Zheng posted 5 hours ago

x^5+x+1=0

Plus and minus x^2 on the LHS of the equation

x^5+x^2-x^2+x+1=0

Group the terms as below

(x^5-x^2)+(x^2+x+1)=0

Factoring the terms in the following steps

x^2(x^3-1)+(x^2+x+1) = 0

x^2(x-1)(x^2+x+1)+(x^2+x+1) = 0

Extract common factor x^2+x+1

(x^2+x+1) (x^2-x^2+1) = 0

Now we have converted the fifth degree equation to the product of two factors. Hence, the following two equations are obtained.

x^2-x^2+1 = 0
(1)
x^2+x+1=0
(2)

From the first equation, we get

x^3-x^2 = -1

Solving the second equation gives two complex roots of the quintic equation

x_1 = -\dfrac{1}{2}+\dfrac{\sqrt{3} }{2}i = ω

x_2 = -\dfrac{1}{2}-\dfrac{\sqrt{3} }{2}i =\overline{ω}

which are conjugate cube root of unities.

Therefore, we have the following properties for the conjugate cube root of unities.

ω^2 = \overline{ω}

ω\overline{ω} = 1

ω^3 = ω

If x = ω,

x^3-x^2 = ω^3-ω^2 = ω-1

If x = \overline{ω}

x^3-x^2 = \overline{ω}^3-\overline{ω}^2 =\overline{ω}-1

Therefore, x^3-x^2 has three solutions.

\begin{cases} -1 \\ ω-1 \\ \overline{ω}-1 \end{cases}

Steven Zheng posted 4 hours ago

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