can anyone solve (999)^1/3 by using binomial theorem?
plz don't write the ans directly.....solve it.
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can anyone solve (999)^1/3 by using binomial theorem?
plz don't write the ans directly.....solve it.
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We know ,according to binomial expansion ,
( x ± a)ⁿ
= xⁿ ( 1 + a/x)ⁿ
= xⁿ( 1 + na/x)
where x >>a
now,
( 999)⅓ = ( 1000 - 1)⅓
= (1000)⅓{ 1 - 1/1000)⅓
= (10)³×⅓ { 1 - 1/1000)⅓
= 10 { 1 - 1/1000}⅓
we know , 1 >> 1/1000
hence,
= 10( 1 - 1/1000)⅓ = 10{1 -1/3 × 1/1000}
= 10 { 1 - 1/3000}
= 10 × 0.99666
=9.9666