Given, p(x) = f(x³) + xg(x³)

We know, x² + x + 1 = (x $$-$$ $$\omega$$) (x $$-$$ $$\omega$$²)

Given, p(x) is divisible by x² + x + 1. So, roots of p(x) is $$\omega$$ and $$\omega$$².

As root satisfy the equation,

So, put x = $$\omega$$

p($$\omega$$) = f($$\omega$$³) + $$\omega$$g($$\omega$$³) = 0

= f(1) + $$\omega$$g(1) = 0 [$$\omega$$³ = 1]

= f(1) + $$\left( { - {1 \over 2} + {{i\sqrt 3 } \over 2}} \right)$$ g(1) = 0

$$ \Rightarrow $$ f(1) $$-$$ $${{g(1)} \over 2} + i\left( {{{\sqrt 3 g(1)} \over 2}} \right)$$ = 0 + i0

Comparing both sides, we get

f(1) $$-$$ $${{g(1)} \over 2}$$ = 0

and $${{{\sqrt 3 } \over 2}g(1) = 0}$$ $$ \Rightarrow $$ g(1) = 0

So, f(1) = 0

Now, p(1) = f(1) + 1 . g(1) = 0 + 0 = 0