Show that if $x^2+x+1=0$, then $x^{26}+\dfrac{1}{x^{26}}=-1$

Question

I am trying to answer the following precalculus question:

Show that if $x^2+x+1=0$, then $A=x^{26}+\dfrac{1}{x^{26}}=-1$.

Let's multiply the first equality (that we know is true) by $(x-1)$. We can do that as $x$ is obviously $\ne1$, $x=1$ isn't a solution of the first equation. (We need the second equality to hold for all of the solutions of the equation). We arrive at $$(x-1)(x^2+x+1)=0\iff x^3-1=0\iff x^3=1\iff x=1, x\in\mathbb{R}$$ in real numbers (I am not familiar with complex numbers), but we said $x\ne 1$. So $x=1$ isn't a solution, so maybe for the left both complex roots the second equality holds.

How can we use that to show that $A=-1$?

Answer 1

You proved correctly that $x^2+x+1=0\implies x^3=1$. But then $x^{26}=x^{27-1}=\frac{(x^3)^9}x=\frac1x$and $\frac1{x^{26}}=x$. So, $$ x^{26}+\frac1{x^{26}}=\frac1x+x=\frac{1+x^2}x=\frac{-x}x=-1. $$

Answer 2

It could be solved with simple applications of the quadratic formula and using the De-Moivre's principle for complex numbers.

To start, let's solve the quadratic equation $x^2 + x + 1 = 0$ using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

In this case, $a = 1$, $b = 1$, and $c = 1$, so:

$$ x = \frac{-1 \pm \sqrt{1 - 4(1)(1)}}{2(1)} $$

Calculating the discriminant ($b^2 - 4ac$):

$$ \text{Discriminant} = 1 - 4(1)(1) = -3 $$

Since the discriminant is negative, the solutions for $x$ are complex:

$$ x = \frac{-1 \pm i\sqrt{3}}{2} $$

Now, we have two values for $x$:

1. $x_1 = \frac{-1 + i\sqrt{3}}{2}$
2. $x_2 = \frac{-1 - i\sqrt{3}}{2}$

Next, we want to find $x^{26}$ and $\frac{1}{x^{26}}$ for these values. Using De Moivre's Theorem, which states:

$$ z^n = r^n \left(\cos(n\theta) + i\sin(n\theta)\right) $$

where $z = r(\cos \theta + i\sin \theta)$, we can calculate $x_1^{26}$ and $x_2^{26}$ as follows:

1. $x_1^{26} = \left(\frac{1}{2}\right)^{26} \left(\cos\left(-\frac{26\pi}{3}\right) + i\sin\left(-\frac{26\pi}{3}\right)\right)$
2. $x_2^{26} = \left(\frac{1}{2}\right)^{26} \left(\cos\left(\frac{26\pi}{3}\right) + i\sin\left(\frac{26\pi}{3}\right)\right)$

Calculating the cosine and sine values:

$$ \cos\left(-\frac{26\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} $$ $$ \sin\left(-\frac{26\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} $$

$$ \cos\left(\frac{26\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} $$ $$ \sin\left(\frac{26\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} $$

Now, we have:

1. $x_1^{26} = 2^{26} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$
2. $x_2^{26} = 2^{26} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$

Both $x_1^{26}$ and $x_2^{26}$ are equal to $2^{26} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$. Now, let's find their reciprocals:

1. $\frac{1}{x_1^{26}} = 2^{26} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$
2. $\frac{1}{x_2^{26}} = 2^{26} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$

Both reciprocals are equal to $2^{26} \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)$. Hence, we have shown that $A = x^{26} + \frac{1}{x^{26}} = -1$, as desired.

Answer 3

Let $P(x)=x^2+x+1$ and $Q(x)=(x-1)P(x)=x^3-1.$

The two polynomials

• $x^{26}-x^2=x^2((x^3)^8-1)=x^2\sum_{k=1}^8\binom8kQ(x)^k$
• $x^{52}-x=x((x^3)^{17}-1)=x\sum_{k=1}^{17}\binom{17}kQ(x)^k$

are multiples of $Q(x).$

Therefore, the polynomial $x^{52}+x^{26}+1=x^{52}-x+x^{26}-x^2+P(x)$ is a multiple of $P(x).$

Answer 4

Another approach:

The given equation $x^2+x+1=0$ can be solved using the quadratic formula (with $a=b=c=1$), obtaining:

$$x = \frac{-1 \pm i\sqrt{3}}{2}$$

Since there are two solutions, I'll distinguish them with subscripts.

$$x_1 = \frac{-1 + i\sqrt{3}}{2}; x_2 = \frac{-1 - i\sqrt{3}}{2}$$

The hard part is calculating $x^{26}$. You could use De Moivre's formula, as @CarnotEngine did, but since you describe yourself as “not familiar with complex numbers”, you probably haven't learned it yet. So let's do this without trig (though we will still need basic arithmetic with complex numbers).

Recall that $(a + ib)^2 = a^2 + 2iab + i^2b^2 = (a^2 - b^2) + 2iab$. From this, we obtain:

$$x_1^2 = \frac{-1 - i\sqrt{3}}{2} = x_2$$ $$x_2^2 = \frac{-1 + i\sqrt{3}}{2} = x_1$$

So, squaring either root produces the other root. That's interesting. Let's use this fact as a shortcut for computing $x^3$.

$$x_1^3 = x_1^2x_1 = x_2x_1$$ $$x_2^3 = x_2^2x_2 = x_1x_2$$

So $x_1^3 = x_2^3 = x_1x_2 = \left( \frac{-1 + i\sqrt{3}}{2} \right)\left( \frac{-1 - i\sqrt{3}}{2}\right) = \frac{1 - 3i^2}{4} = \frac{4}{4} = 1$. In general, multiplying two complex conjugates together results in a real number. We now have $x^3 = 1$, the same intermediate result from the OP.

Raising both sides to the 8th power gives $x^{24} = 1$. We can use this to help calculate $x^{26}$.

$$x_1^{26} = x_1^{24}x_1^2 = x_1^2 = x_2 = \frac{-1 - i\sqrt{3}}{2}$$ $$x_2^{26} = x_2^{24}x_2^2 = x_2^2 = x_1 = \frac{-1 + i\sqrt{3}}{2}$$

For the reciprocal, $\frac{1}{x^{26}}$, let's rationalize the denominators.

$$\frac{1}{x_1^{26}} = \frac{2}{-1 - i\sqrt{3}} \times \frac{-1 + i\sqrt{3}}{-1 + i\sqrt{3}} = \frac{-2 + 2i\sqrt{3}}{4} = \frac{-1 + i\sqrt{3}}{2} = x_1$$ $$\frac{1}{x_2^{26}} = \frac{2}{-1 + i\sqrt{3}} \times \frac{-1 - i\sqrt{3}}{-1 - i\sqrt{3}} = \frac{-2 - 2i\sqrt{3}}{4} = \frac{-1 - i\sqrt{3}}{2} = x_2$$

Now we can evaluate $x^{26} + \frac{1}{x^{26}}$.

$$x_1^{26} + \frac{1}{x_1^{26}} = x_2 + x_1$$ $$x_2^{26} + \frac{1}{x_2^{26}} = x_1 + x_2$$

Conveniently, you get the same value whether you start with $x_1$ or $x_2$. Just add the two roots. The imaginary parts cancel out and you get:

$$\boxed{x^{26} + \frac{1}{x^{26}} = -1}$$

QED