If ω≠1 is a cube root of unity and ω+xn=1+12ω+69ω+ then the values of n and x - Maths - Binomial Theorem

Question

I f   ω ≠ 1   i s   a   c u b e
  r o o t   o f   u n i t y   a n d  
ω + x n = 1 + 12 ω + 69 ω + t h e n   t h e
  v a l u e s   o f   n   a n d   x  
a r e   r e s p e c t i v e l y ( A ) 36 , 1 ( B ) 12 , 2 ( C
) 24 , 1 / 2 ( D ) 18 , 1 / 3

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

Third term should be 69ω2By binomial theoremω+xn=nC0ωn+nC1xωn-1+nC2x2ωn-2+
=ωn+n xωn-1+nn-12!x2ωn-2+
On comparing it with 1+12ω+69ω2+
ωn=1 in xωn-1=12ω iinn-12!x2ωn-2=69ω2
iiiiii/ii2nn-12!x2ωn-2n2 x2ω2n-2=69ω2144ω2nn-1ωnω22n2 ω2nω2=69144n-1ωn2n ω2n=69144Use ωn=1n-12n=69144⇒144n-144=138n⇒6n=144⇒n=24Put in ii24 xω23=12ω⇒x=12ω22⇒x=12ω21
ω⇒x=12ω As ω3k=1, where k is integerHence none of the options are correct

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I f ω ≠ 1 i s a c u b e r o o t o f u n i t y a n d ω + x n = 1 + 12 ω + 69 ω + . . . . t h e n t h e v a l u e s o f n a n d x a r e r e s p e c t i v e l y ( A ) 36 , 1 ( B ) 12 , 2 ( C ) 24 , 1 / 2 ( D ) 18 , 1 / 3

$I f ω \neq 1 i s a c u b e r o o t o f u n i t y a n d {(ω + x)}^{n} = 1 + 12 ω + 69 ω + . . . . t h e n t h e v a l u e s o f n a n d x a r e r e s p e c t i v e l y (A) 36, 1 (B) 12, 2 (C) 24, 1 / 2 (D) 18, 1 / 3$