Question

ix)prove that (a + b)2 + (aw + bw2)2 + (aw2+ bw)2=6ab
​

Answer 1

Answer:

Proved

Step-by-step explanation:

a + b)² + (aω + bω²)² + (aω² + bω)² = 6ab

Step-by-step explanation:

(a+b)^2+(aw+bw^2)^2+(aw^2+bw)^2=6ab

LHS =

(a + b)² + (aω + bω²)² + (aω² + bω)²

= a² + b² + 2ab + a²ω² + b²ω⁴ + 2abω³ + a²ω⁴ + b²ω² + 2abω³

= a² + b² + 2ab + a²ω² + b²ω³.ω + 2abω³ + a²ω³.ω + b²ω² + 2abω³

ω³ = 1

= a² + b² + 2ab + a²ω² + b²ω + 2ab + a²ω + b²ω² + 2ab

= a² + a²ω² + a²ω + b² + b²ω + b²ω² + 2ab + 2ab + 2ab

= a²(1 + ω² + ω) + b²(1 + ω + ω²) + 6ab

1 + ω + ω² = 0

= 0 + 0 + 6ab

= 6ab

= RHS

QED

Proved

(a + b)² + (aω + bω²)² + (aω² + bω)² = 6ab

Answer 2

Answer:

We know that

${w}^{3} = 1 \: \: \: and \: \: 1 + w + {w}^{2} = 0$

Hence the given expression

$= {(a + b)}^{2} + {(aw + b {w}^{2} )}^{2} + {(a {w}^{2} + bw)}^{2}$

$= {a}^{2} + {b}^{2} + ab+ a^{2} {w}^{2} + {b}^{2} {w}^{4} +2ab {w}^{3} + {a}^{2} {w}^{4} + {b}^{2} {w}^{2} + 2ab {w}^{3}$

$= {a}^{2} + {b}^{2} + ab+ a^{2} {w}^{2} + {b}^{2} {w} +2ab + {a}^{2} {w} + {b}^{2} {w}^{2} + 2ab$

$= {a}^{2} (1 + w + {w}^{2} )+ {b}^{2} (1+ w + {w}^{2} ) + 6ab$

$= 6ab$