Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

According to my book

In the given equation $$x^2+x+1=0\tag{1}$$

If $a$ is a root of eqn $(1)$ then $a$ satisfies the following equation

$$a^2+a+1=0\tag{2}$$

$$\implies (a-1)(a^2+a+1)=0\tag{3}$$

$$a^3=1$$

How do you get the last $3$ equations? Also, how does $(3)$ follow from $(2)$? It's not the same thing mathematically.

EDIT: Okay I've got my answer, although I still have another doubt. How do we get $a^2+a+1=0$ merely because $a$ is a root of the original equation?

share|improve this question
    
Equation $(3)$ is weaker than equation $(2)$. You obtain it by multiplying $(2)$ with $(a-1)$. –  Daniel Fischer Sep 4 '13 at 15:00
    
It's like if $a=1\implies a^2=1\implies a=\pm1$ –  lab bhattacharjee Sep 4 '13 at 15:00
    
Thanks. But how do we get equation(2) just because a is a root of $x^2+x+1$ ? I've never come across this in all my time dealing with quadratic equation. –  Ghost Sep 4 '13 at 15:04
    
@Ghost, it is the definition of root. –  njguliyev Sep 4 '13 at 15:25

2 Answers 2

up vote 1 down vote accepted

We say that $a$ is a root of $f(x)=x^2+x+1$ if $f(a)=a^2+a+1=0$. Hence equation $(2)$ follows form $(1)$, if we say that $a$ is a root. $(3)$ follows from $(2)$ since $0=0\cdot (a-1)=(a^2+a+1)(a-1)=a^3-1$.

share|improve this answer

Well, if $\;z=0\;$ then clearly $\,w\cdot z=0\;$ for any $\;w\;$ , right? Here it is the same, after you open up parentheses:

$$a^2+a+1=0\implies (a-1)(a^2+a+1)=0 .\;\text{But}\;\;$$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.