Question

Solve the attached question.... urgent....​

Answer 1

Step-by-step explanation:

We have,

$x = a + b \omega + c { \omega}^{2} \\ y = a \omega + b { \omega}^{2} + c \\ z = a { \omega}^{2} + b + c\omega$

Now,

$x + y + z = a(1 + \omega + \omega^{2} ) + b(1 + \omega + \omega^{2} ) +c (1 + \omega +\omega^{2} )$

We know, $1+\omega+\omega^2=0$

So,

$\implies \: x + y + z = a(0) + b(0 ) +c (0 )$

$\implies \: x + y + z = 0$

$\implies \: x + y = - z$

$\implies \: (x + y)^{3} = ( - z) ^{3}$

$\implies \: x^{3} + y^{3} + 3 {x}^{2} y + 3x {y}^{2} = - z ^{3}$

$\implies \: x^{3} + y^{3} + 3x y (x+ y) + z ^{3} = 0$

$\implies \: x^{3} + y^{3} + 3x y ( - z) + z ^{3} = 0$

$\implies \: x^{3} + y^{3} - 3x y z + z ^{3} = 0$

$\implies \: x^{3} + y^{3} + z ^{3} = 3xyz$

$\implies \: \frac{ x^{3} + y^{3} + z ^{3}}{xyz} = \frac{ 3xyz}{xyz}$

$\implies \: \frac{ x^{3} }{xyz} +\frac{ y^{3} }{xyz} + \frac{ z^{3} }{xyz} = 3$

$\implies \: \frac{ x^{2} }{yz} +\frac{ y^{2} }{zx} + \frac{ z^{2} }{xy} = 3$