Cube Roots

The Original Method

An inspection of this table reveals that each cube ends in a different digit. This
digit corresponds to the cube root in all cases except 2 and 3 and 7 and 8. In
these four cases the final digit of the cube is the difference between the cube root
and 10.

Number Cube
1 1
2 8
3 27
4 64
5 125
6 216
7 343
8 512
9 729
10 1000

To see how this information is used, let us suppose
that a person calls out the cube 250047.

Step 1: The last number is a 7 which tells us immediately that the last number of the cube root must be 3.

Step 2: The first
number of the cube root is determined as follows. Discard the last three figures
of the cube (regardless of the size of the number) and consider the remaining
figures—in this example they are 250. In the above table, 250 lies between the
cubes of 6 and 7. The lower of the two figures—in this case 6—will be the first
figure of the cube root. The correct answer, therefore, is 63.

The alternate technique

My method involves presenting a different perspective to the first step of finding the last digit. For people already comfortable with the original method, this might not seem appealing as it involves a shift in their thinking and a relearning of a classical method. But for new players, this could be a chance to show that they are different.

Last Digit(D) (0/5 + $x^3$) mod 10
1 1
8 8
7 27
4 64
5 125
6 216
3 343
2 512
9 729

We introduce $x$ where { $x < 5$}. Based on the above table, we generate a variable $0+x$ or $5+x$. Depending on the location of D in the table, we choose which variable to use and after we fill in the value of $x$, that variable becomes our last digit of our root.

But how do we obtain $x$? We solve (0 + $x^3$) mod 10 or (5 + $x^3$) mod 10 (depending on where D is in the table) for $x$ such that
(0 + $x^3$) mod 10 = D for all D in {1,8,7,4}
(5 + $x^3$) mod 10 = D for all D in {5,6,3,2,9}

Example 1.) $77^3=456533$

Step 1: Last digit of our cube is 3. 3 is generated by solving (5+ $2^3$) mod 10 where 2 is the value of $x$. So we obtain our variable 5+x = 5 + 2 = 7. This is our last digit.

Step 2: As in the usual method, after slashing the last 3 digits, we obtain 456 which is greater than 343 which is cube of 7. Thus the first digit is 7.

Hence, we arrive at our cube root of 77.

Example 2.) \$49^3=117,649\$

Step 1: Last digit of our cube is 9. 9 is generated by solving (5+ $4^3$) mod 10 where 4 is the value of $x$. So we obtain our variable 5+x = 5 + 4 = 9. This is our last digit.

Step 2: As in the usual method, after slashing the last 3 digits, we obtain 117 which is greater than 64 which is cube of 4. Thus the first digit is 4.

Hence, we arrive at our cube root of 49.

Example 3.) $63^3=250,047$

Step 1: Last digit of our cube is 7. 7 is generated by solving (0+ $3^3$) mod 10 where 3 is the value of $x$. So we obtain our variable 0+x = 0 + 3 = 3. This is our last digit.

Step 2: As in the usual method, after slashing the last 3 digits, we obtain 250 which is greater than 216 which is cube of 6. Thus the first digit is 6.

Hence, we arrive at our cube root of 63.