Tough question if x^2+x+1=0 then find x^2018+1/x^2018
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Given,
x²+ x +1 = 0
Use quadratic formula,
x =(-1 ±i√3)/2
x = -1/2 ± i√3/2
x = cos(180-60) ± isin(180-60)
x = cos2π/3 ±isin2π/3
Let x = cos2π/3 + isin2π/3
and 1/x = cos2π/3 -isin2π/3
Now,
x^2018 + 1/x^2018
= (Cos2π/3 +i sin2π/3)^2018 +( cos2π/3 -i.cos2π/3)^2018
Use De- Moivre's theorem
=(Cos2π×2018/3+ i.sin2π×2018/3)+(cos2π×2018/3-i.sin2π×2018/3)
=2cos(4036π/3)
=2cos(2×1345π/2 + π/3)
=2cos(2 ×(672 + 1/2)π + π/3)
= 2cos(2×672π + π+ π/3)
= 2cos(π + π/3)
= - 2cos(π/3)
= -2 × 1/2 = -1
x²+ x +1 = 0
Use quadratic formula,
x =(-1 ±i√3)/2
x = -1/2 ± i√3/2
x = cos(180-60) ± isin(180-60)
x = cos2π/3 ±isin2π/3
Let x = cos2π/3 + isin2π/3
and 1/x = cos2π/3 -isin2π/3
Now,
x^2018 + 1/x^2018
= (Cos2π/3 +i sin2π/3)^2018 +( cos2π/3 -i.cos2π/3)^2018
Use De- Moivre's theorem
=(Cos2π×2018/3+ i.sin2π×2018/3)+(cos2π×2018/3-i.sin2π×2018/3)
=2cos(4036π/3)
=2cos(2×1345π/2 + π/3)
=2cos(2 ×(672 + 1/2)π + π/3)
= 2cos(2×672π + π+ π/3)
= 2cos(π + π/3)
= - 2cos(π/3)
= -2 × 1/2 = -1
farhan123:
u made it hard it is very easy
I GAVE REAL CONCEPT , I KNOW THAT
abhi i m in 9 but i know trignometry i m in kota in resonance
the simple ans. is if x^2 +x+1=0 then x^3=1 and now divide x^2+x+1=0 whole eq. by x then x+1/x=-1 and then x^2.x^2016+1/x^2.x^2016 = x^2 +1/x^2 and last ans is -1..its that simple only applying common sense in paper of about 1 n a half hour
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