Question

if x=a+b,y=aw+bw2,z=aw2+bw show that x3+y3+z3=3(a3+b3)​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \blue{x = a + b}} \\ \: \: \: \: \: \: \: \: \tt{ \blue{y = a \omega + b \omega^{2} }} \\ \: \: \: \: \: \: \: \tt{ \blue{z = a \omega^{2} + b \omega}}$

Now,

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

$\sf{ \implies \: x + y + z = (a + a \omega +a \omega^{2}) +( b + b \omega + b \omega^{2} ) }$

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

$\sf{ \implies \: x + y + z =a (0) +b( 0 ) }$

$\sf{ \implies \: x + y + z = 0 }$

So,

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

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3(a + b)(a \omega + b \omega^{2} )(a \omega^{2} + b \omega) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3(a ^{2} \omega + ab \omega^{2} +ab \omega + {b}^{2} \omega^{2} )(a \omega^{2} + b \omega) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 \{a ^{2} \omega + ab (\omega^{2} + \omega) + {b}^{2} \omega^{2} \}(a \omega^{2} + b \omega) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 \{a ^{2} \omega + ab ( - 1) + {b}^{2} \omega^{2} \}(a \omega^{2} + b \omega) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 (a ^{2} \omega - ab + {b}^{2} \omega^{2} )(a \omega^{2} + b \omega) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 (a ^{3} \omega ^{3} - a^{2} b \omega^{2} + a{b}^{2} \omega^{4} + {a}^{2}b \omega^{2} - ab^{2} \omega + {b}^{3} \omega^{3} ) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 (a ^{3} - a^{2} b \omega^{2} + a{b}^{2} \omega^{3} \cdot \omega + {a}^{2}b \omega ^{2} - ab^{2} \omega + {b}^{3} ) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 (a ^{3} - a^{2} b \omega^{2} + a{b}^{2}\omega + {a}^{2}b \omega ^{2} - ab^{2} \omega + {b}^{3} ) }$

$\sf{ \implies \: x^{3} + y^{3} + z^{3} = 3 (a ^{3} + {b}^{3} ) }$