If $a^2 + a + 1 = 0$ find $a^3$

Question

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

This is how I had solved the question by using the identity :- $$a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$$

But the roots of the equation in question are complex :- $$x = \frac{-1}{2} \pm \frac{\sqrt 3i}{2}$$

I am not able to understand how this works.

Answer 1

I do believe the instructor posing this question wanted the student to observe that the polynomial $a^3-1$ factors into $$a^3-1 = (a-1)(a^2+a+1).$$ So if $a$ satisfies $a^2+a+1 = 0$ then $a$ must also satisfy $a^3-1=0$ or equivalently, $a^3=1$.

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A slightly different way: From the equation $a^2+a+1=0$ one may derive the equation $$a(a^2+a+1) = a^3+a^2+a = a \cdot 0 = 0,$$ or in particular, $$a^3+a^2+a=0.$$ However, subtracting each side of the equation $a^2+a+1=0$ from the equation $a^3+a^2+a = 0$ gives $$a^3+a^2+a - (a^2+a+1) = a^3-1 = 0.$$ So adding $1$ to each side of this gives $a^3=1$.

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There is, in the posing of this question, something special about the exponent $3$ here. If the teacher had asked for the value of $a^2$ instead of $a^3$ say, then there could be two possible values, one for each root of $a^2+a+1$.

Answer 2

Another way $a^2=-a-1$

$a^3=a^2 a=(-a-1)a=-a^2-a=a+1-a=1$

Answer 3

• a²+a+1=0
• Divide by "a"
• a+1+1/a=0
• a²+a+1=a+1+1/a
• a²=1/a
• a³=1

But, It's quite contradictory If, a³=1 a=1

• 1+1+1≠0

for, a²+a+1=0 Or a(a²+a+1)=0

a³+a²+a=0 a²+a=-1 So, that's true... Trick is that a³=1, but a≠1... a=-a²-1 Complex

Answer 4

The discriminant of $a^2+a+1$ is negative and so you know the solutions will be complex non-real. As you need complex solutions for $a$, you should thus write it as a complex number: \begin{equation} a=r[\cos(\theta) + i \sin (\theta)]. \end{equation} Insert this in the equation \begin{equation} a^3-1=0 \end{equation} and find the solutions for $r$ and $\theta$. This yields: \begin{align} a^3-1&=0\\ \Leftrightarrow r^3 [\cos(\theta) + i \sin (\theta)]^3 - 1 &= 0\\ \Leftrightarrow r^3 [\cos(3\theta) + i \sin (3\theta)] &= 1 \end{align} (the development of the power in the last equality stemming from the formula of De Moivre).

Now the moduli and arguments must match on each side, so you have \begin{equation} \begin{cases} r^3 = 1\\ 3\theta = 2 \pi k \end{cases} \end{equation} (with $k \in \mathbb{Z}$), and hence \begin{equation} \begin{cases} r = 1\\ \theta = \displaystyle\frac{2 \pi k}{3} \end{cases}. \end{equation}

The three principal solutions (i.e. with $\theta \in [0, 2 \pi[$) of $a$ are thus given (in polar form) by \begin{align} a &= 1 \cdot [\cos(0) + i \sin (0)] = 1\\ a &= 1 \cdot [\cos(\frac{2 \pi}{3}) + i \sin (\frac{2 \pi}{3})]\\ a &= 1 \cdot [\cos(\frac{4 \pi}{3}) + i \sin (\frac{4 \pi}{3})]\\ \end{align} The first solution, $a=1$, is there because of the factor $(a-1)$ you added to your original equation and should be discarded. If you write the latter two solutions in Cartesian form instead of polar form you obtain the two complex solutions you mentioned.

Answer 5

From $a^2=-a-1$ we get as $a\neq0$ that $a^3=-a^2-a=1$.