Question

can anyone solve (999)^1/3 by using binomial theorem?
plz don't write the ans directly.....solve it.

Answer 1

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

Answer 2

( 999)⅓ = ( 1000 - 1)⅓

= (1000)⅓{ 1 - 1/1000)⅓

= (10)³×⅓ { 1 - 1/1000)⅓

= 10 { 1 - 1/1000)⅓

1 >> 1/1000

= 10( 1 - 1/1000)⅓ = 10{1 -1/3 × 1/1000}

= 10 { 1 - 1/3000}

= 10 × 0.99666

=9.9666