Question

If w be a complex cube root of unity and x= a+b, y= a+bw, z= a+bw², show that
x³+y³+z³=3(a³+b³).
(complex number chapter)

Answer 1

Answer:

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Answer 2

Q. WHY YOU ASK THE SAME QUESTION FOR MULTIPLE TIME?

I RECOMMEND YOU TO SOLVE THESE TYPE OF QUESTION IN PARTS AND ADD THESE AT LAST....

ANSWER....

PART I

${x}^{3} = {(a + b)}^{3} \\ \\ {x}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

PART II

$y = a + bw \\ \\ {y}^{3} = {(a + bw)}^{3} \\ \\ {y}^{3} = {a}^{3} + {(bw)}^{3} + 3abw(a + bw) \\ \\ we \: know \: that \: {w}^{3} = 1 \\ \\ {y}^{3} = {a}^{3} + {b}^{ 3} + w(3 {a}^{2} b + 3a {b}^{2}w)$

PART III

$z = a + b {w}^{2} \\ \\ {z}^{ 3} = {a}^{3} + {b}^{ 3} {w}^{6} + 3ab {w}^{2} (a + b {w}^{2} ) \\ \\ we \: know \: that \: {w}^{6} = 1 \\ \\ {z}^{3} = {a}^{3} + {b}^{3} + {w}^{2} (3 {a}^{2} b + 3a {b}^{2} )$

ADD ALL THESE PARTS...

${x}^{3} + {y}^{3} + {z}^{3} = 3( {a}^{3} + {b}^{3} ) + (3 {a}^{2} b + 3a {b}^{2} )(1 + {w}^{2} + w) \\ \\ and \: i \: again \: write \: most \: important \: \\ formula \: of \: complex \: no.(iit \: level) \\ \\ 1 + {w}^{2} + w = 0 \\ \\ than \: after \: that \: we \: got \\ \\ {x}^{3} + {y}^{3} + {z}^{3} = 3( {a}^{3} + {b}^{3} )$

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