Cube roots of unity: 1, w, $w^2$ where w = (-1+i*sqrt(3))/2 = e^$\frac{2*pi*i}{3}$.

Fundamental Properties:

• $w^3$ = 1 and 1 + w + $w^2$ = 0
• $w^2$ = w-bar (they are conjugates)
• |w| = |$w^2$| = 1
• w^(-1) = $w^2$

Simplification Substitutions:

• 1 + w = -$w^2$ (most used substitution)
• 1 + $w^2$ = -w
• w + $w^2$ = -1

Useful Products:

• (1-w)(1-$w^2$) = 3
• (1+w)(1+$w^2$) = 1
• (a+bw+$cw^2$)(a+$bw^{2+cw}$) = a^{2+b}^{2+c}^{2-ab-bc-ca}

Factorizations:

• $x^{2+x+1}$ = (x-w)(x-$w^2$)
• $x^{3-1}$ = (x-1)(x-w)(x-$w^2$)
• a^{3+b}^{3+c}^{3-3abc} = (a+b+c)(a+bw+$cw^2$)(a+$bw^{2+cw}$)

Power Reduction: For any integer n, $w^n$ = w^(n mod 3). This converts any power of w to one of {1, w, $w^2$}.

JEE Problem Strategy: When you see w in a problem:

1. Immediately write 1+w+$w^2$ = 0
2. Reduce all powers mod 3
3. Replace 1+w with -$w^2$ (or similar)
4. Simplify using $w^3$ = 1

These identities appear in determinants, binomial coefficients, and algebraic expression problems. Master the substitutions to solve in 30-60 seconds.