Question

p+qw+rw²/
r+pw+qw² =w²​

Answer 1

SOLUTION

TO PROVE

$\displaystyle \sf{ \frac{p + q \omega + r { \omega}^{2} }{r + p \omega + q { \omega}^{2} } = { \omega}^{2} \: }$

$\displaystyle \sf{ where \: \omega \: \: is \: cube \: root \: of \: unity }$

PROOF

$\displaystyle \sf{ \because \: \omega \: \: is \: cube \: root \: of \: unity }$

So

$\sf{ { \omega}^{3} = 1 \: }$

Now

$\displaystyle \sf{ \frac{p + q \omega + r { \omega}^{2} }{r + p \omega + q { \omega}^{2} } \: }$

Multiplying numerator and denominator both by $\sf{ { \omega}^{2} \: }$

So

$\displaystyle \sf{ \frac{p + q \omega + r { \omega}^{2} }{r + p \omega + q { \omega}^{2} } \: }$

$= \displaystyle \sf{ \frac{ { \omega}^{2} (p + q \omega + r { \omega}^{2}) }{ { \omega}^{2}(r + p \omega + q { \omega}^{2} )} \: }$

$= \displaystyle \sf{ \frac{ { \omega}^{2} (p + q \omega + r { \omega}^{2}) }{(r { \omega}^{2} + p { \omega}^{3} + q { \omega}^{4} )} \: }$

$= \displaystyle \sf{ \frac{ { \omega}^{2} (p + q \omega + r { \omega}^{2}) }{r { \omega}^{2} +( p \times 1 )+( q \times { \omega}^{} \times { \omega}^{3}) } \: }$

$= \displaystyle \sf{ \frac{ { \omega}^{2} (p + q \omega + r { \omega}^{2}) }{r { \omega}^{2} + p+q { \omega}^{} } \: }$

$= \displaystyle \sf{ \frac{ { \omega}^{2} (p + q \omega + r { \omega}^{2}) }{ (p + q \omega + r { \omega}^{2}) } \: }$

$= \sf{ { \omega}^{2} \: }$

Hence proved

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Answer 2

Step-by-step explanation:

Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit. To students of electronics, Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists. E = I x R.