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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $a + b + c = 0$.

Similar Questions

One of the complex roots of the equation $x^{11}-x^6-x^5+1=0$ is

If $\alpha, \beta$ are the roots of the equation $x^2-4x+8=0$,then for any $n \in N$,$\alpha^{2n}+\beta^{2n}$ equals

${\left[ {\frac{{1 + \cos (\pi /8) + i\sin (\pi /8)}}{{1 + \cos (\pi /8) - i\sin (\pi /8)}}} \right]^8}$ is equal to

$\alpha, \beta$ are the roots of the equation $x^2+2x+4=0$. If the point representing $\alpha$ in the Argand diagram lies in the $2^{nd}$ quadrant and $\alpha^{2024}-\beta^{2024}=ik, (i=\sqrt{-1})$,then $k=$

The general value of the real angle $\theta$, which satisfies the equation $(\cos \theta + i \sin \theta)(\cos 2\theta + i \sin 2\theta) \dots (\cos n\theta + i \sin n\theta) = 1$ is given by (assuming $k$ is an integer):

Vedclass Test Series

Exam Paper Generator

Online Exam Module