Question

root 1+ a2+ root 1 + a2+ a4=​

Answer 1

Answer:

Since, 1,a1,a2,....,an−1 are the nth root of unity.

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)∴xn−1+xn−2+...+x2+x+1

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)∴xn−1+xn−2+...+x2+x+1=(x−a1)(x−a2)....(x−an−1)

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)∴xn−1+xn−2+...+x2+x+1=(x−a1)(x−a2)....(x−an−1)Put x=1, we get

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)∴xn−1+xn−2+...+x2+x+1=(x−a1)(x−a2)....(x−an−1)Put x=1, we get(1−a1)(1−a2)...(1−an−t)=1+1+...+n times

Since, 1,a1,a2,....,an−1 are the nth root of unity.∴xn−1=(x−1)(x−a1)...(x−an−1)⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)∴xn−1+xn−2+...+x2+x+1=(x−a1)(x−a2)....(x−an−1)Put x=1, we get(1−a1)(1−a2)...(1−an−t)=1+1+...+n times=n. 

Step-by-step explanation:

It may help you .

Please mark me as brainlist.

Answer 2

$ANSWER</p><p></p><p>Since, 1,a1,a2,....,an−1 are the nth root of unity.</p><p>∴xn−1=(x−1)(x−a1)...(x−an−1)</p><p>⇒x−1xn−1=(x−a1)(x−a2)....(x−an−1)</p><p>∴xn−1+xn−2+...+x2+x+1</p><p>=(x−a1)(x−a2)....(x−an−1)</p><p>Put x=1, we get</p><p>(1−a1)(1−a2)...(1−an−t)=1+1+...+n times</p><p>=n. </p><p></p><p>$

this is your answer dear plz mark as