Given that
\[
\sum_{k=1}^{n} k(k-1)=\frac{n(n-1)(n+1)}{3}
\]
and $\omega$ and $\omega^2$ are complex cube roots of unity. If
\[
\sum_{k=1}^{2026}
\left(
k+\frac{1}{\omega}
\right)
\left(
k+\frac{1}{\omega^2}
\right)
=
\frac{2026}{3}(N+3),
\]
then $N=$
Show Hint
For cube roots of unity:
\[
1+\omega+\omega^2=0
\]
is the most important identity and is frequently used in simplification problems.
Step 1: Use the cube root facts. For cube roots of unity, $1+\omega+\omega^2=0$ and $\omega^3=1$. Also $\dfrac{1}{\omega}=\omega^2$ and $\dfrac{1}{\omega^2}=\omega$. Step 2: Simplify the product inside the sum. So $\left(k+\dfrac1\omega\right)\left(k+\dfrac1{\omega^2}\right)=(k+\omega^2)(k+\omega)$. Open it: $k^2+k(\omega+\omega^2)+\omega^3$. Step 3: Replace the known sums. Since $\omega+\omega^2=-1$ and $\omega^3=1$, this equals $k^2-k+1$. So each term is simply $k^2-k+1=k(k-1)+1$. Step 4: Add over all $k$. The full sum is $\displaystyle\sum_{k=1}^{2026}\big(k(k-1)+1\big)=\sum k(k-1)+\sum 1$. Step 5: Use the given identity. Here $\sum_{k=1}^{n}k(k-1)=\dfrac{n(n-1)(n+1)}{3}$ with $n=2026$, giving $\dfrac{2026\cdot2025\cdot2027}{3}$, and $\sum 1=2026$. Step 6: Take out the common factor and match. Factor $\dfrac{2026}{3}$: the total is $\dfrac{2026}{3}\big(2025\times2027+3\big)$. Comparing with $\dfrac{2026}{3}(N+3)$ gives $N=2025\times2027=2027\times2025$. \[ \boxed{2027\times2025} \]