Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal
Disclaimer:
1 b) $(2 + \omega + \omega^2)^3 + (1 + \omega - \omega^2)^8 - (1 - 3 \omega + \omega^2)^4 = 1$
Solution:
We know,
$$1 + \omega + \omega^2 = 0$$
LHS
$= (2 + \omega + \omega^2)^3 + (1 + \omega - \omega^2)^8 - (1 - 3 \omega + \omega^2)^4$
$= (1 + 1 + \omega + \omega^2)^3 + ( - \omega ^2 - \omega^2)^8 - ( - \omega - 3 \omega)^4$
$= (0+1)^3 + (-2 \omega^2)^8 - (-4 \omega)^4$
$= 1 + 2^8 \omega^{16} - 2^8 \omega^4$
$= 1 + 2^8 \omega^4 ( \omega ^12 - 1)$
$= 1 + 2^8 \omega^3 \omega ( (\omega^3)^4 - 1)$
$= 1 + 2^8 (1) \omega ( (1)^4 - 1)$
$= 1 + 2^8 \omega (1 - 1)$
$= 1 + 2^8 \omega \times 0$
$= 1$
RHS
1 c) $(1 - \omega + \omega^2)^4 (1 + \omega - \omega^2) ^4 = 256
Solution:
LHS
$= (1 - \omega + \omega^2)^4 (1 + \omega - \omega^2) ^4$
$= (- \omega - \omega)^4 (-\omega^2 - \omega^2)^4$
$= (-2 \omega)^4 (-2\omega^2)^4$
$= \{ (-2 \omega)(-2 \omega^2) \}^4$
$= \{ 4 \omega^3 \}^4$
$= 4^3 (\omega^3)^4$
$= 256 \times 1$
$= 256$
RHS
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