Evaluate (1 056)^1/3 correct up to the four places of decimal - Maths - Binomial Theorem

Question

Evaluate (1 056)^1/3 correct up to the four places of decimal

Ishwarmani · Accepted Answer

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

We have,1 05613=1+0
05613As, 1+xn=1+nx+nn-12!x2+nn-1n-23!x3+ So, 1 05613=1+0
05613=1+130 056+13-23120 0562+ =1+0 01867-190 003136+ =1 01867-0
00035=1 01832=1 0183   Approx

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Evaluate (1.056)^1/3 correct up to the four places of decimal.