# Confusion with qudratic equations

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?

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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

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 $0=0\cdot (a-1)=(a^2+a+1)(a-1)=a^3-1$.
Well, if $\;z=0\;$ then clearly $\,w\cdot z=0\;$ for any $\;w\;$ , right? Here it is the same, after you open up parentheses:
$$a^2+a+1=0\implies (a-1)(a^2+a+1)=0 .\;\text{But}\;\;$$
$$(a-1)(a^2+a+1)=a^3-1\implies a^3-1=0\iff a^3=1$$