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प्रश्न
Use binomial theorem to evaluate the following upto four places of decimal
`root(3)(126)`
उत्तर
`root(3)(126) = (126)^(1/3)`
= `(125 + 1)^(1/3)`
= `[125(1 + 1/125)]^(1/3)`
= `(125)^(1/3) (1 + 1/125)^(1/3)`
= `5[1 + 1/3(1/125) + (1/3(1/3 - 1))/(2!) (1/125)^2 + ...]`
= `5[1 + 1/375 + 1/3(-2/3)(1/2)(1/15625) + ...]`
= 5 [1 + 0.00266 – 0.000007 + ... ]
= 5 (1.00266) = 5.0133, upto 4 places of decimals.
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