The coefficient of $$x^{256}$$ in the expansion of $$(1-x)^{101}(x^2 + x + 1)^{100}$$ is:

We need the coefficient of $$x^{256}$$ in $$(1-x)^{101}(x^2+x+1)^{100}$$.

Observe the factorisation $$1 - x^3 = (1-x)(x^2+x+1)$$, so $$(x^2+x+1)^{100} = \frac{(1-x^3)^{100}}{(1-x)^{100}}$$.

Therefore the expression becomes: $$(1-x)^{101} \cdot \frac{(1-x^3)^{100}}{(1-x)^{100}} = (1-x)(1-x^3)^{100}.$$

We need the coefficient of $$x^{256}$$ in $$(1-x)(1-x^3)^{100}$$, which equals $$[\text{coeff of }x^{256}\text{ in }(1-x^3)^{100}] - [\text{coeff of }x^{255}\text{ in }(1-x^3)^{100}].$$

The general term in $$(1-x^3)^{100}$$ is $$\binom{100}{k}(-1)^k x^{3k}$$, so only powers that are multiples of 3 appear.

Since $$256$$ is not a multiple of $$3$$, the coefficient of $$x^{256}$$ in $$(1-x^3)^{100}$$ is $$0$$.

Since $$255 = 3 \times 85$$, the coefficient of $$x^{255}$$ in $$(1-x^3)^{100}$$ is $$\binom{100}{85}(-1)^{85} = -\binom{100}{85} = -\binom{100}{15}$$.

Therefore the coefficient of $$x^{256}$$ in the original expression is $$0 - \left(-\binom{100}{15}\right) = \binom{100}{15}.$$