Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal
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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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