Question

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

Answer 1

$(126{)}^{\frac{1}{5}}=(5^3+1{)}^{\frac{1}{3}}$
$=5{(1+\frac{1}{5^3})}^{\frac{1}{3}}$
$=5(1+\frac{1}{3}\cdot \frac{1}{5^3}-\frac{1}{9}\cdot \frac{1}{5^6}+\frac{5}{81}\cdot \frac{1}{5^9}-....)$
$=5+\frac{1}{3}\cdot \frac{1}{5^2}-\frac{1}{9}\cdot \frac{1}{5^5}+\frac{5}{81}\cdot \frac{1}{5^7}-....$
$=5+\frac{1}{3}\cdot \frac{2^2}{{10}^2}-\frac{1}{9}\cdot \frac{2^5}{{10}^5}+\frac{1}{81}\cdot \frac{2^7}{{10}^7}-....$
$=5+\frac{\cdot 04}{3}-\frac{\cdot 00032}{9}+\frac{\cdot 0000128}{81}-...$
$5+\cdot 0133333...-\cdot 000035...+...$
$=5\cdot 01329,$ to five places of decimals.