Solve - a - b - - c -2c - a - - b -2 -

Given: a+bω+cω^2c+aω+bω^2×ω^2ω^2=ω^2(a+bω+cω^2)cω^2+aω^3+bω^4 =ω^2(a+bω+cω^2)cω^2+a+bω #160; (since, ω^3=1 ) =ω^2 ...

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Standard XII

Mathematics

nth Term of A.P

Solve : a +...

Question

Solve : $\frac{(a + b ω + c ω^{2})}{c + a ω + b ω^{2}} =$ ___

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Solution

Given: $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}} \times \frac{ω^{2}}{ω^{2}} = \frac{ω^{2} (a + b ω + c ω^{2})}{c ω^{2} + a ω^{3} + b ω^{4}}$
$= \frac{ω^{2} (a + b ω + c ω^{2})}{c ω^{2} + a + b ω}$ (since, $ω^{3} = 1$ )
$= ω^{2}$

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(\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}) + (\frac{a + b ω + c ω^{2}}{b + c ω + a ω^{2}})

\frac{(a + b ω + c ω^{2})}{c + a ω + b ω^{2}} = ω^{2}

P r o v e t h a t (\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}) = ω^{2}, w h e r e ω i s t h e c u b e r o o t o f u n i t y

Q. The value of

\frac{a + b ω + c ω^{2}}{b + c ω + a ω^{2}} + \frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}

will be

If $ω$ is a complex cube root of unity, then the value of $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}$ + $\frac{a + b ω + c ω^{2}}{b + c ω + b ω^{2}}$ is :

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Solve : (a+bω+cω2)c+aω+bω2=___

Given: a+bω+cω2c+aω+bω2×ω2ω2=ω2(a+bω+cω2)cω2+aω3+bω4=ω2(a+bω+cω2)cω2+a+bω (since, ω3=1)=ω2

Solve : $\frac{(a + b ω + c ω^{2})}{c + a ω + b ω^{2}} =$ ___

Given: $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}} \times \frac{ω^{2}}{ω^{2}} = \frac{ω^{2} (a + b ω + c ω^{2})}{c ω^{2} + a ω^{3} + b ω^{4}}$
$= \frac{ω^{2} (a + b ω + c ω^{2})}{c ω^{2} + a + b ω}$ (since, $ω^{3} = 1$ )
$= ω^{2}$