Let =-1+i√32 and =-1-i√32,i=-1.

Question

Let $\alpha =\frac{-1+i\sqrt{3}}{2}$ and $\beta =\frac{-1-i\sqrt{3}}{2},i=\sqrt{-1}$. If

$(7-7\alpha +9\beta {)}^{20}+(9+7\alpha -7\beta {)}^{20}+(-7+9\alpha +7\beta {)}^{20}+(14+7\alpha +7\beta {)}^{20}={m}^{10},$

then $m$ is $____$

Answer: 49

Accepted Answer

Here $\alpha,\beta$ are the two non-real cube roots of unity, i.e. $\alpha=\omega,\ \beta=\omega^2$, so $\alpha+\beta=-1$ and $\alpha\beta = 1$.

Since $\alpha+\beta=-1$, the fourth term simplifies immediately:
$$14+7\alpha+7\beta = 14+7(\alpha+\beta)=14-7=7 \implies (14+7\alpha+7\beta)^{20}=7^{20}.$$

For the first three terms, each is (up to a rotation by a power of $\omega$) a multiple of the quantity $w=(-7+9\alpha+7\beta)$; carrying out the substitution shows
$$(7-7\alpha+9\beta)^{20}+(9+7\alpha-7\beta)^{20}+(-7+9\alpha+7\beta)^{20} = \big(1+\alpha^{20}+\beta^{20}\big)\,w^{20}.$$

Since $20 \equiv 2 \pmod 3$, $1+\alpha^{20}+\beta^{20}=1+\omega^{2}+\omega^{4}=1+\omega^2+\omega=0$ (using $1+\omega+\omega^2=0$). Hence the first three terms sum to $0$, leaving

$$m^{10}=0+7^{20} \implies m^{10}=(7^2)^{10} \implies m=49.$$

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