Calculate the following expressions for α = cos(2π)/(3) +…
Calculate the following expressions for $\alpha = \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}$: $$ 1^{\circ}\ \ (\alpha a+\alpha^2 b)(\alpha^2 a+\alpha b); \qquad 2^{\circ}\ \ (a+b+c)(a+\alpha b+\alpha^2 c)(a+\alpha^2 b+\alpha c). $$
Explanation of Method
$\alpha$ is a primitive cube root of unity, so two identities drive the whole computation: $\alpha^3=1$ (so any power of $\alpha$ reduces mod 3), and $1+\alpha+\alpha^2=0$ (so $\alpha+\alpha^2=-1$, a real number, even though $\alpha$ and $\alpha^2$ are individually complex). Expanding each product and repeatedly applying these two facts collapses everything down to real, symmetric expressions in $a$, $b$, $c$.
Step 1 · Identify $\alpha$
By Euler's formula, $\alpha = e^{i2\pi/3} = \cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3} = -\dfrac12+i\dfrac{\sqrt3}{2}$.
Step 2 · Key properties of $\alpha$
Since $\alpha=e^{i2\pi/3}$: $$ \alpha^3 = e^{i2\pi} = 1 $$ so $\alpha$ is a primitive cube root of unity. Because $\alpha$ satisfies $z^3-1=(z-1)(z^2+z+1)=0$ and $\alpha\neq1$, it must satisfy the second factor:
$$ \boxed{1+\alpha+\alpha^2 = 0} \qquad\Longrightarrow\qquad \alpha+\alpha^2=-1 $$
Also, $\alpha^2 = e^{i4\pi/3} = -\dfrac12-i\dfrac{\sqrt3}2 = \bar\alpha$, and $\alpha^4=\alpha^3\cdot\alpha=\alpha$.
Part 1° · $(\alpha a+\alpha^2 b)(\alpha^2 a+\alpha b)$
Expanding:
$$ (\alpha a+\alpha^2 b)(\alpha^2 a+\alpha b) = \alpha^3a^2 + \alpha^2 ab + \alpha^4 ab + \alpha^3 b^2 $$
Using $\alpha^3=1$ and $\alpha^4=\alpha$:
$$ = a^2+b^2+ab(\alpha+\alpha^2) $$
and substituting $\alpha+\alpha^2=-1$:
$$ \boxed{(\alpha a+\alpha^2 b)(\alpha^2 a+\alpha b) = a^2-ab+b^2} $$
Part 2° · $(a+b+c)(a+\alpha b+\alpha^2 c)(a+\alpha^2 b+\alpha c)$
Write $P_0=a+b+c$, $P_1=a+\alpha b+\alpha^2c$, $P_2=a+\alpha^2b+\alpha c$ — these are $P(1), P(\alpha), P(\alpha^2)$ for $P(x)=a+xb+x^2c$.
Step 2a · Compute $P_1 \cdot P_2$
Expanding term by term and using $\alpha^3=1$, $\alpha^4=\alpha$:
$$ P_1P_2 = a^2+b^2+c^2 + ab(\alpha+\alpha^2) + ac(\alpha+\alpha^2) + bc(\alpha+\alpha^2) $$
Substituting $\alpha+\alpha^2=-1$:
$$ P_1P_2 = a^2+b^2+c^2-ab-ac-bc $$
Step 2b · Multiply by $P_0$
$$ P_0P_1P_2 = (a+b+c)(a^2+b^2+c^2-ab-ac-bc) $$
Expanding fully, the mixed terms like $ab^2, a^2b, ac^2, a^2c, bc^2, b^2c$ each appear once with $+$ and once with $-$ and cancel, leaving:
$$ \boxed{(a+b+c)(a+\alpha b+\alpha^2 c)(a+\alpha^2 b+\alpha c) = a^3+b^3+c^3-3abc} $$
This matches the well-known factorization identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, confirming the result.
The three cube roots of unity $1, \alpha, \alpha^2$, spaced $120°$ apart on the unit circle, summing to $0$.
Result
For $\alpha = \cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}$ (a primitive cube root of unity, satisfying $\alpha^3=1$ and $1+\alpha+\alpha^2=0$):
Both results are real (and symmetric or antisymmetric in $a,b,c$) despite involving the complex number $\alpha$ throughout — a direct consequence of $\alpha+\alpha^2=-1$ always being real.