Cunning complex 3

$(a+b)(a^2-ab+b^2)=1001$ ................... $(1)$

$a^2 \omega^{3} +ab(\omega+\omega^{2})+b^2 \omega^{3}$

we know that $\omega^{3}=1~and ~1+\omega^{2}+\omega=0$

expression reduces to $a^2 -ab+b^2$

as in $1$

$a+b=1$ , thus $a^2 -ab+b^2=1001$

$(a{ \omega }^{ 2 }+b\omega )(a\omega +b{ \omega }^{ 2 })={ a }^{ 2 }+{ b }^{ 2 }+ab(\omega +{ \omega }^{ 2 })={ a }^{ 2 }+{ b }^{ 2 }-ab.\quad \\ \frac { { a }^{ 3 }+{ b }^{ 3 } }{ a+b } ={ a }^{ 2 }+{ b }^{ 2 }-ab=\frac { 1001 }{ 1 } =1001.$

Since w^3 = 1 and w is complex and imaginary, w^2 + w + 1 = 0.

(aw^2 + bw)(aw + bw^2) = (a^2)(w^3) + ab(w^4) + ab(w^2) + (b^2)(w^3) = (a^2 + b^2) + abw + ab(w^2) = (a^2 + b^2) + ab(w + w^2) = a^2 + b^2 - ab

From the systems for a and b, a^2 - ab + b^2 = 1001.

Hence, the answer.

Multiplying and dividing by w. We get (aw+b)(aw^2+b)= a^2+b^2+ab (w+w^2). This is same as (a+b)^2 -3ab=1-3(-1000/3)= 1001