x+1/x=square roots of3 then the find the value of x^18+x^12+x^6+1

Question

we want solution of  above problem.                                                                                 and it is from quadratic function chapter.

Answer 1

If (x+1)/x=√3, then x²-x√3+1=0. Therefore from the quadratic formula, x=(√3±√(3-4))/2=√3/2±i/2 where i=√-1. As a complex number, x can be written x=e^±⅙πi=cos(π/6)±isin(π/6), because cos(π/6)=√3/2, sin(π/6)=1/2.

x¹⁸=e^±3πi=cos(±3π)±isin(3π)=-1; x¹²=e^2πi=1; x⁶=e^πi=-1. Cosine is an even function (cos(nπ)=cos(-nπ)), and sin(nπ)=0 for all integer n.

Therefore x¹⁸+x¹²+x⁶+1=-1+1-1+1=0.

Note:

if x+1/x=-√3 (the "other" square root of 3), x=-√3/2±i/2=cos(5π/6)±isin(5π/6), but x¹⁸=x⁶=-1, as before, because cos(±15π)=cos(±3π)=cos(±5π)=cos(±π)=-1; and x¹²=1, because cos(±10π)=cos(±2π)=cos(0)=1 and sin(nπ)=0, so the result is the same.