HOW TO FIND THE CUBE ROOT OF A NUMBER WHICH IS NOT A PERFECT CUBE - Maths - Cubes and Cube Roots

Question

HOW TO FIND THE CUBE ROOT OF A NUMBER WHICH IS NOT A PERFECT
CUBE ?

Ajanta Trivedi · Accepted Answer

we can find the cube root of a number by long division method,

it is similar to the long division method of finding the square root.

but for finding the cube root we will make the set of 3 numbers.

let us consider one example of finding the cube root of 12812905.

here we will make the groups of three numbers starting from the unit digit going left and after decimal point we will make the groups of three numbers going right.

so 12812905 = 12, 812, 905

the left most group is 12. so we will take the number whose cube is less than 12.

$2 2 \sqrt[3]{\bar{12} \bar{812} \bar{905}} \underset{̲}{8}____4 812$

the digit 2 thus obtained is our $a_{n}$ . next we will try to find the largest digit $a_{n - 1}$ so that $(30 a_{n} \cdot a_{n} a_{n - 1} + (a_{n - 1})^{2}) a_{n - 1}$ does not exceed 4812.

thus $a_{n - 1} = 3$

ie. $(30 * 2 * 23 + 3^{2}) * 3 = 1389 * 3 = 4167$ whereas $(30 * 2 * 24 + 4^{2}) * 4 = 5824$ which is greater than 4812.

$2 3 2 \sqrt[3]{\bar{12} \bar{812} \bar{905}} \underset{̲}{8}____4 812 \underset{̲}{4167}___645 905$

next we will try to find $a_{n - 2}$ . such that $(30 * 23 * 23 a_{n - 2} + (a_{n - 2})^{2}) a_{n - 2}$ does not exceed 645905.

here $23 a_{n - 2}$ is a 3 digit number where unit digit is $a_{n - 2}$ . and hundred's digit is 2 and ten's digit is 3.

thus we find $a_{n - 2}$ = 4.

since $(30 * 23 * 234 + 4^{2}) * 4 = 645904$

$2 34.001 2 \sqrt[3]{\bar{12} \bar{812} \bar{905} . \bar{000} \bar{000}} \underset{̲}{8}____4 812 \underset{̲}{4167}___645 905 \underset{̲}{645904}_____1 000 000 000$

similarly we can find the cube root of other numbers and till the required accuracy.

hope this helps you.

cheers!!

HOW TO FIND THE CUBE ROOT OF A NUMBER WHICH IS NOT A PERFECT CUBE ?