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प्रश्न
Find `root(3)(10001)` approximately (two decimal places
उत्तर
`root(3)(10001) = (10001)^(1/3)`
= `(1000 + 1)^(1/3)`
= `{1000(1 + 1/1000)}^(1/3)`
= `(1000)^(1/3) [1 + 1/10^3]^(1/3)`
= `10{1 + 1/3(1/10^3) + (1/3((-2)/3))/3 (1/10^3)^2 ...}`
= `10{1 + 1/3000 - 2/18000000 ...}`
= 10[1 + 0.000333 ...]
= 10(1.000333)
= 10.0033
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