Advertisements
Advertisements
Question
If ω ≠ 1 is a cube root of unity, show that (1 – ω + ω2)6 + (1 + ω – ω2)6 = 128
Advertisements
Solution
ω is a cube root of unity ω3 = 1, 1 + ω + ω2 = 0
(1 – ω + ω2)6 + (1 + ω – ω2)6
= (– ω – ω)6 + (– ω2 – ω2)6
= (– 2ω)6 + (– 2ω2)6
= (– 2)6(ω6 + ω12)
= (64)(1 + 1)
= 128
APPEARS IN
RELATED QUESTIONS
If to ω ≠ 1 is a cube root of unity, then show that `("a" + "b"omega + "c"omega^2)/("b" + "c"omega + "a"omega^2) + ("a" + "b"omega + "c"omega^2)/("c" + "a"omega + "a"omega^2)` = – 1
If 2 cos α = `x + 1/x` and 2 cos β = `y + 1/y`, show that `x/y + y/x = 2cos(alpha − beta)`
If 2 cos α = `x + 1/x` and 2 cos β = `y + 1/y`, show that `xy - 1/xy = 2"i" sin(alpha + beta)`
If 2 cos α = `x + 1/x` and 2 cos β = `y + 1/y`, show that `x^"m" y^"n" + 1/(x^"m" y^"n")` = 2 cos(mα – nβ)
If 2cos α = `x + 1/x` and 2 cos β = `y + 1/x`, show that `x^"m"/y^"n" - y^"n"/x^"m"` = 2i sin(mα – nβ)
Solve the equation z3 + 27 = 0
If ω ≠ 1 is a cube root of unity, show that the roots of the equation (z – 1)3 + 8 = 0 are – 1, 1 – 2ω, 1 – 2ω2
Find the value of `sum_("k" = 1)^8 (cos (2"k"pi)/9 + "i" sin (2"kpi)/9)`
If ω ≠ 1 is a cube root of unity, show that (1 + ω)(1 + ω2)(1 + ω4)(1 + ω8)….. (1 + ω2n) = 1
If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when θ = `pi/3`
If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when θ = `(2pi)/3`
If z = 2 – 2i, find the rotation of z by θ radians in the counterclockwise direction about the origin when θ = `(3pi)/3`
Choose the correct alternative:
If ω ≠ 1 is a cubic root of unity and `|(1, 1, 1),(1, - omega^2 - 1, omega^2),(1, omega^2, omega^7)|` = 3k, then k is equal to
Choose the correct alternative:
If ω = `cis (2pi)/3`, then the number of distinct roots of `|(z + 1, omega, omega^2),(omega, z + omega^2, 1),(omega^2, 1, z + omega)|` = 0
