Hint: Look at the options

We know that $x^3=1$ has $3$ roots, they're $1,\omega , \omega ^2$ .

Because of this, we have $1+\omega +\omega ^2=0$ and $\omega ^{3k} = 1$ and from these two, we have $\omega ^{3k} + \omega^{3n+1} + \omega^{3m+2} = 0$

From the options, you're expected to think of the cube roots of unity, so put $x=\omega$ in $f(x) = x^{101} + x^{57} + x^{94} + x^{33} -1$ , and you get

$f(\omega) = \omega ^{101} + \omega ^{57}+\omega ^{94} +\omega ^{33}-1\\ \quad\quad = \omega ^2+1+\omega +1-1 \\\quad \quad = \omega ^2 + \omega +1 \\ \quad\quad = 0$

$f(\omega) =0 \implies \omega$ is a root of $f(x)$ $\implies \omega ^2$ is also a root. (Because $\omega, \omega^2$ are complex conjugates)

Hence we conclude that $\boxed{x^2+x+1}$ is a factor of given polynomial.

The answer must divide evenly into the original equation for ALL values of x. When x=3 we have x^2+x-1=11; x^2-x+1=7; x^2-x-1=5. The powers of 3 repeat the last 2 digits in a pattern mod 20. Using that fact, x^101 + x^94 + x^47 + x^33 - 1 boils down to 3 + 69 + 63 + 23 - 1 = 157 when we sum the last 2 digits when x=3. Since 5 or 7 do not divide evenly into 57, and 11 cannot divide evenly into ...57, the only answer left is x^2 + x + 1.

Substitute 1 in the question and also in the options.Which one of the options matches with the answer obtained in putting one ,that is the answer

x^101+x^94+x^57+x^33-1=(x^2+x+1) (x^99-x^98+x^96-x^95+x^93-x^91+x^90-x^88+x^87-x^85+x^84-x^82+x^81-x^79+x^78-x^76+x^75-x^73+x^72-x^70+x^69-x^67+x^66-x^64+x^63-x^61+x^60-x^58+x^57+x^31-x^30+x^28-x^27+x^25-x^24+x^22-x^21+x^19-x^18+x^16-x^15+x^13-x^12+x^10-x^9+x^7-x^6+x^4-x^3+x-1) = 0

Among

-0.5+0.866025403784439i

0.618033988749895

1.61803398874989

0.5+0.866025403784439i

Only the first one of x^2 + x + 1 = 0 can satisfy it.