If 1, omega, omega^2 are the three cube roots | vedclass

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(A) Given $a + b + c = 0$,we have $c = -(a + b)$.Substitute $c$ into the expression:$E = (a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$Usi

Vedclass · Accepted Answer

$27abc$

Vedclass · Answer

(A) Given $a + b + c = 0$,we have $c = -(a + b)$.Substitute $c$ into the expression:$E = (a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$Using $1 + \omega + \omega^2 = 0$,we know $\omega^2 = -1 - \omega$ and $\omega = -1 - \omega^2$.Alternatively,using the property that for $a+b+c=0$,the expression simplifies to $27abc$.Let $a=1, b=1, c=-2$. Then $a+b+c=0$.Expression $= (1 + \omega - 2\omega^2)^3 + (1 + \omega^2 - 2\omega)^3$$= (1 + \omega - 2(-1 - \omega))^3 + (1 + \omega^2 - 2(-1 - \omega^2))^3$$= (1 + \omega + 2 + 2\omega)^3 + (1 + \omega^2 + 2 + 2\omega^2)^3$$= (3 + 3\omega)^3 + (3 + 3\omega^2)^3$$= 27(1 + \omega)^3 + 27(1 + \omega^2)^3$$= 27(-\omega^2)^3 + 27(-\omega)^3$$= 27(-\omega^6) + 27(-\omega^3)$$= 27(-1) + 27(-1) = -54$.Checking option $A$: $27abc = 27(1)(1)(-2) = -54$.Thus,the expression is equal to $27abc$.

If $1, \omega, \omega^2$ are the three cube roots of unity,then the value of $(a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3$ is equal to,given that $a + b + c = 0$.

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