Question

Find the cube root of 126 to 5 places of decimals.___

Answer 1

$(126{)}^{\frac{1}{3}}=(5^3+1{)}^{\frac{1}{3}}=5{(1+\frac{1}{5^3})}^{\frac{1}{3}}=5(1+\frac{1}{3}.\frac{1}{5^3}-\frac{1}{9}.\frac{1}{5^6}+\frac{5}{81}.\frac{1}{5^9}-\cdots )=5+\frac{1}{3}.\frac{1}{5^2}-\frac{1}{9}.\frac{1}{5^5}+\frac{1}{81}.\frac{1}{5^7}-\cdots =5+\frac{1}{3}.\frac{2^2}{{10}^2}-\frac{1}{9}.\frac{2^5}{{10}^5}+\frac{1}{81}.\frac{2^7}{{10}^7}-\cdots =5+\frac{0.04}{3}-\frac{0.00032}{9}+\frac{0.0000128}{81}-\cdots$
= 5 + 0.013333 . . . - 0.000035 . . . + . . .
= 5.01329, to five places of decimals