Let w be a solution of x²+x+1 = 0 What is the value of w^10 + w^5 + 3?
w is\[\frac{i\sqrt{3}}{2}-1/2.\] Is there a simple way to calculate w^10 and w5 other than doing lots of multiplications?
Oh, and it can be its conjugate too.
substitute p= w^5
All right... Rewriting it, p² + p + 3. But what I asked was how can I make the multiplication not too tiring?
solve for p and you will have two solutions equal to w^2, square them and substitute in equation of w
Well the cube root of 1 is w, and there is a equation that says 1+w+w^2=0 So, from here x= w Now W^10=(w^2)^5= (-(1+w))^5 From here you might be able to calculate.
w^5 *
@kutabs , wouldn't it be that w³ is 1?
Oh right, I got it. The answer is 2, isn't it? THanks for the help.
@Chillout :Of course. w^3=1 Or, w^3-1=0 Or, (w-1)(w^2+w+1)=0 (From a^3-b^3) Either, the roots are w=1 or 1+w+w^2=0
Yeah, 2. :)
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