Question

if w is a complex cue root of unity,then show that (1-w+w^2)^5+(1+w-w^2)^5=32​

Answer 1

Answer:

32

Step-by-step explanation:

To prove--->

( 1 - w + w² )⁵ + ( 1 + w - w² )⁵ = 32

Proof---> First we discuss something about w.

w is cube root of unity , which are three in number and as follows

1 , (-1 + √3i ) / 2 , ( -1 - √3i ) / 2

we represent last two roots as

w = (-1 + √3i ) / 2 , w² = (-1 - √3i ) / 2

Now we have two formulee related to cube root of unity

1 + w + w² = 0

We get three other relations from it as follows

1 + w² = - w , 1 + w = - w² , w + w² = -1

Other formula is as follows

w³ = 1

Now returning to original problem and taking LHS

LHS =(1 - w + w² )⁵ + ( 1 + w - w² )⁵

=( 1 + w² - w )⁵ + ( 1 + w - w² )⁵

Using 1 + w² = -w and 1 + w = -w² , we get

= ( - w - w )⁵ + ( - w² - w² )⁵

= ( -2 w )⁵ + ( -2w² )⁵

Using law of exponent

( a b c )ᵐ = aᵐ b ᵐ cᵐ ,we get

= (-1)⁵ (2)⁵ w⁵ + (-1)⁵ ( 2 )⁵ ( w² )⁵

If (-1) has odd exponent than its value is (-1) and using this aᵐ⁺ⁿ = aᵐ aⁿ we get

= - 1 (32) w³ w² + (-1) (32) w¹⁰

Using w³ = 1

= -32 (1) w² - 32 w⁹ w

Using w³ = 1 again

= -32 w² - 32 ( w³ )³ w

= -32 w² - 32 (1)³ w

= -32 w² - 32 w

= -32 ( w + w² )

Now using w + w² = -1

= -32 ( -1 )

= 32 = RHS

Answer 2

Answer:

Step-by-step explanation: