If x=a+b+c, y=a α+b β+c and z=a β+b α+c, where α and β are complex cube roots of unity, then x y z=

Question

If x=a+b+c, y=a α+b β+c and z=a β+b α+c, where α and β are complex cube roots of unity, then x y z= (A) 2(a3+b3+c3) (B) 2(a3-b3-c3) (C) a3+b3+c3-3 a b c (D) a3-b3-c3

Tardigrade Admin · Accepted Answer

x=a+b+c, y=a ω+b ω2+c, z=a ω2+b ω+c y z=(a ω+b ω2+c)(a ω2+b ω+c) =a2+b2+c2+a b(ω4+ω2)+b c(ω2+ω)+c a(ω2+ω) =a2+b2+c2+a b(ω2+ω)+b c(ω2+ω)+c a(ω2+ω) =a2+b2+c2-a b-b c-c a ω2+ω+1=0 x y z=(a+b+c)(a2+b2+c2-a b-b c-c a) =a3+b3+c3-3 a b c

Q. If $x = a + b + c, y = a α + b β + c$ and $z = a β + b α + c$ , where $α$ and $β$ are complex cube roots of unity, then $x yz =$

$2 (a^{3} + b^{3} + c^{3})$

$2 (a^{3} - b^{3} - c^{3})$

$a^{3} + b^{3} + c^{3} - 3 ab c$

$a^{3} - b^{3} - c^{3}$

Q. If x=a+b+c,y=aα+bβ+c and z=aβ+bα+c, where α and β are complex cube roots of unity, then xyz=

2(a3+b3+c3)

2(a3−b3−c3)

a3+b3+c3−3abc

a3−b3−c3

x=a+b+c,y=aω+bω2+c,z=aω2+bω+c yz=(aω+bω2+c)(aω2+bω+c) =a2+b2+c2+ab(ω4+ω2)+bc(ω2+ω)+ca(ω2+ω) =a2+b2+c2+ab(ω2+ω)+bc(ω2+ω)+ca(ω2+ω) =a2+b2+c2−ab−bc−ca{ω2+ω+1=0} xyz=(a+b+c)(a2+b2+c2−ab−bc−ca) =a3+b3+c3−3abc

Q. If $x = a + b + c, y = a α + b β + c$ and $z = a β + b α + c$ , where $α$ and $β$ are complex cube roots of unity, then $x yz =$

$2 (a^{3} + b^{3} + c^{3})$

$2 (a^{3} - b^{3} - c^{3})$

$a^{3} + b^{3} + c^{3} - 3 ab c$

$a^{3} - b^{3} - c^{3}$