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Put $$a=2025$$. The divisor is $$a^2+a+1$$, and $$(a-1)(a^2+a+1)=a^3-1$$, so $$a^3\equiv1$$ modulo the divisor. Since $$2026\equiv1\pmod3$$, we get $$a^{2026}\equiv a$$, and hence the required remainder is $$0$$.
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