Question

1+w+w^2 =0 how ...
Just explain with brief

Answer 1

Briefly: 
1, w and w² are the 3 cube-roots of 1.
Or they are the roots of the polynomial x³ - 1=0

In the polynomial x³-1=0; a=1, b=0,c=0, d = -1
Sum of roots =1+w+w² = -b/a = 0/1 = 0.

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1, w and w² are the 3 cube-roots of 1. 
They are the roots of the equation x³ - 1 = 0

x³ - 1 = 0
⇒ (x-1)(x²+x+1) = 0
Thus x-1 = 0 or x=1
And x²+x+1 = 0

⇒ x = $\frac{-1+i \sqrt{3} }{2} \ and\ \frac{-1-i \sqrt{3} }{2}$

So three roots are $1,\ \frac{-1+i \sqrt{3} }{2} \ and\ \frac{-1-i \sqrt{3} }{2}$

$Let\ w=\frac{-1+i \sqrt{3} }{2} \\ then\ w^2= \frac{-1-i \sqrt{3} }{2}$

$1+w+w^2=1+\frac{-1+i \sqrt{3} }{2}+ \frac{-1-i \sqrt{3} }{2} =\frac{2-1-1+i \sqrt{3}-i \sqrt{3} }{2}=0$