State true or false- - is an imaginary root of unity- a - b- - c-23 - a - b-2 - c-3 - 2a -b-c2b -a-c2c-a-b-

#160; a + bω+cω^2 = x and a + bω^2 + cω=y . Then, #160; (a+bω +cω^2)^3 + (a+bω^2+cω)^3 = x^3 + y^3 = (x+y)(x+ω y)(x+ ω^2y) Now, #160; x ...

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Byju's Answer

Standard XII

Mathematics

Inequality

State true or...

Question

State true or false:
$ω$ is an imaginary root of unity.
$(a + b ω + c ω^{2})^{3} + (a + b ω^{2} + c ω)^{3} = (2 a - b - c) (2 b - a - c) (2 c - a - b)$ .

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Solution

$a + b ω + c ω^{2} = x$ and $a + b ω^{2} + c ω = y$ . Then,
$(a + b ω + c ω^{2})^{3} + (a + b ω^{2} + c ω)^{3} = x^{3} + y^{3} = (x + y) (x + ω y) (x + ω^{2} y)$
Now,
$x + y = (a + b ω + c ω^{2}) + (a + b ω^{2} + c ω)$
$= 2 a + b (ω + ω^{2}) + c (ω + ω^{2})$
$= 2 a - b - c$
$x + ω y = (a + b ω + c ω^{2}) + ω (a + b ω^{2} + c ω)$ $= (1 + ω) a + (1 + ω) b + 2 ω^{2} c$
$= ω^{2} (2 c - a - b)$
Similarly,
$x + ω^{2} y = ω (2 b - a - c)$
$⟹ (x + y) (x + b ω y) (x + ω^{2} y)$
$= ω^{3} (2 a - b - c) (2 c - a - b) (2 b - a - c)$
$= (2 a - b - c) (2 c - a - b) (2 b - a - c)$

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Similar questions

ω

is an imaginary root of unity.

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(a + b ω + c ω^{2})^{3} + (a + b ω + c ω^{2} + c ω)^{3} = 27 a b c

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Q. If cube root of unity is

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State true or false:ω is an imaginary root of unity. (a+bω+cω2)3+(a+bω2+cω)3=(2a−b−c)(2b−a−c)(2c−a−b).

State true or false:
$ω$ is an imaginary root of unity.
$(a + b ω + c ω^{2})^{3} + (a + b ω^{2} + c ω)^{3} = (2 a - b - c) (2 b - a - c) (2 c - a - b)$ .