Roots of a quadratic equation

Question

If $ a $ , $b$ are the roots of $ x^2+x+1=0 $ , then the equation whose roots are $a^k $ and $b ^k$ where k is a positive integer not divisible by 3 is

$a)$ $x^2 - x + 1 = 0$

$b)$ $x^2 + x+1 = 0$

$c)$ $ x^2 -x -1 =0$

$d)$ None of the above

My attempt : I was able to solve the question , only after I used the options given below . I put k = 2 and checked the options for which the above statement was coming to be true.Using the formula for $(a+b)^2$ .

However I am looking for a more genera approach , preferably one which doesn't apply checking against the options

Answer 1

Hint: the roots are $$\omega,\omega^2$$ so sum of roots of the wanted quadratic is $\omega^k+\omega^{2k}=(-1)$ as $3|k$ (excluded) and product is equal to $\omega^{2k+k}=1$ thus the correct option is $x^2+x+1$.

Answer 2

Since $x^3-1=(x-1)(x^2+x+1)$, you have $a^3=1$ and $b^3=1$.

If $k=3h+1$, then $a^k=a(a^3)^h=a$ and $b^k=b(b^3)h=b$.

On the other hand, $a^2=b$ and $b^2=a$, so if $k=3h+2$, you have $a^k=a^2(a^3)^h=b$ and $b^k=b^2(b^3)^h=a$.

Why is $a^2=b$? It is clear that $(a^2)^3=(a^3)^2=1$, so $a^2$ is a root of $x^3-1$. However, $a^2\ne1$ because neither $1$ nor $-1$ are roots of $x^2+x+1$; it cannot be $a^2=a$, because otherwise $a=0$ or $a=1$. So the only remaining possibility is $a^2=b$. For the same reason, $b^2=a$.