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Discriminant

Given that ...

Question

Given that $1, ω, ω^{2}$ are cube roots of unity, show that $(1 - ω + ω^{2})^{5} + (1 + ω - ω^{2})^{5} = 32$ .

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Solution

Given,

$(1 - w + w^{2})^{5} + (1 + w - w^{2})^{5}$

$1 + w + w^{2} = 0, w^{3} = 1$

$= (- w - w)^{5} + (- w^{2} - w^{2})^{5}$

$= (- 2 w)^{5} + (- 2 w^{2})^{5}$

$= (- 2)^{5} (w + w^{2})^{5}$

$= (- 32) (- 1)^{5}$

$= (- 32) (- 1)$

$= 32$

Hence proved.

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Q. The value of

{(1 - ω + ω^{2})}^{5} + {(1 + ω - ω^{2})}^{5}

ω

ω^{2}

are the complex cube roots of unity, is

1, ω, ω^{2}

are the three cube roots of unity, find the value of

{(1 - ω + ω^{2})}^{5} + {(1 + ω - ω^{2})}^{5}

.

ω

is a cube root of unity, then

(1 + ω - ω^{2})^{5} + (1 - ω + ω^{2})^{5} =

ω

is a non-real cube root of unity, then the value of

{(1 - ω + ω^{2})}^{5} + {(1 + ω - ω^{2})}^{5}

1, ω, ω^{2}

are the cube roots of unity, then prove that

\frac{1}{2 + ω} + \frac{1}{1 + 2 ω} = 1 + ω^{2}

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