If $$\omega \, (\neq 1)$$ is a cube root of unity, and $$(1 + \omega)^7 = A + B\omega$$. Then $$(A, B)$$ equals:

For the cube roots of unity we know

$$\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0.$$

From $$1 + \omega + \omega^2 = 0$$ we obtain

$$1 + \omega = -\omega^2.$$

Now evaluate powers of $$1+\omega$$.

Step 1: Cube.
$$(1+\omega)^3 = (-\omega^2)^3 = -\,\omega^{6}.$$

Because $$\omega^3 = 1 \implies \omega^{6} = (\omega^3)^2 = 1,$$ we get

$$(1+\omega)^3 = -1.$$

Step 2: Sixth power.
$$(1+\omega)^6 = \bigl((1+\omega)^3\bigr)^2 = (-1)^2 = 1.$$

Step 3: Seventh power.
$$(1+\omega)^7 = (1+\omega)^6\,(1+\omega) = 1\,(1+\omega) = 1+\omega.$$

Thus

$$(1+\omega)^7 = 1 + \omega = A + B\omega,$$

which gives $$A = 1,\; B = 1.$$

Therefore $$(A, B) = (1, 1).$$

Option A which is: $$(1, 1)$$