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Question

If ` ω ` is cube root of unity and x+y+z=ax+ωy+ω2 z =b, x +ω2y+ωz=c then which of the following is correct

A
x=a+b+c3
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B
y=a+bω2+ωc3
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C
Z=a+bω+ω2c3
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D
All the above
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Solution

The correct option is C Z=a+bω+ω2c3
1+ω+ω2=0 and ω3=1
x+y+z=a
x+wy+w2z=b
+x+w2y+wz=c
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯3x+(1+ω+ω2)y+(1+ω+ω2)z=a+b+c

3x=a+b+cx=a+b+c3

y+z=a(a+b+c3)=2abc3

y=2abc3z3

a+b+c3+w(2abc3z)3+w2z=b

a+b+c+2awbwcw3wz+3w2z=3b2b
a2b+c+2awbwcw=3wz3w2z
a(1+w)+awb(1+w)b+c(1w)=3wz(1w)
aw2+aw+bw2b+c(1w)=3wz(1w)
aw(1w)b(1w)(1+w)+c(1w)=3wz(1w)
aw+bw2+c=3wz

z=w2(aw+bw2+c)3w3=a+bw+cw23
y=2abcabwcw23
y=ab(1+w)c(1+w2)3
y=a+bw2+cw3

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