If N is number and n is another number such that N = n^{3}, we say n is the cube root of N and write n = cube root of N .
i.e.,
Examples for Cube root
Example 12: Find the cube root of 216 by factorization.
Solution:
216 = 2 x (108) = 2 x 2 x 54 = 2 x 2 x 2 x 27 = 2 x 2 x 2 x 3 x 9 = 2 x 2 x 2 x 3 x 3 x 3
216 = (2 x 3) x (2 x 3) x (2 x 3)
216 = 6 x 6 x 6 = 6^{3}
Example 13: find the cube root of 17576 using factorization.
Solution:
17576 = 2 x (8788) = 2 x 2 x (4394) = 2 x 2 x 2 x (2197) = 2 x 2 x 2 x (13) x (13) x (13)
= [2 x (13)] x [2 x (13)] x [2 x (13)]
= (26) x (26) x (26)
= (26)^{3}
Example 15: Find the cube root of 103823.
Solution:
Here unit in the digits place is 3. If n^{3} = 103823. Then, the unit in the digits place of n must be 7.
Let us split this as 103 and 823.
We observe that, 4^{3 }= 64 < 103 < 125 = 5^{3}.
Hence 40^{3} = 64000 < 103823 < 125000 = 50^{3}. Hence n must lie between 40 and 50. Since the unit in the digits place is 7, therefore n must be 47.
47^{3} = 103823.
Cube root – Chapter 2 – Exercise 1.2.8

Find the cube root by prime factorization.
i) 10648
Solution:
10648 = 2 x 5324
= 2 x 2 x 2662
= 2 x 2 x 2 x 1331
= 2 x 2 x 2 x 11 x 121
= 2 x 2 x 2 x 11 x 11 x 11
= (2 x 11) x (2 x 11) x (2 x 11)
= 22 x 22 x 22
= 22^{3}.
ii) 46656
Solution:
46656 = 2 x (23328)
= 2 x 2x (11664)
= 2 x 2 x 2 x (5832)
= 2x2x2x2x(2916)
= 2x2x2x2x2x(1458)
= 2x2x2x2x2x2x(729)
= 2x2x2x2x2x2x9x(81) = 2x2x2x2x2x2x9x9x9 = (2x2x9)x(2x2x9)x(2x2x9) = 36 x 36 x36
= 36^{3}.
iii)15625
Solution:
15625 = 5 x (3125)
= 5 x 5 x (625)
= 5 x 5 x 5 x (125)
= 5 x 5 x 5 x 5 x (25)
= 5 x 5 x 5 x 5 x 5 x 5
= (5×5)x(5×5)x(5×5)
= (25) x (25) x (25)
= 25^{3}

Find the cube root of the following by looking at the last digit and using estimation.
i) 91125
Solution:
Here unit in the digits place is 5. If n^{3} = 91125. Then, the unit in the digits place of n must be 5.
Let us split this as 91 and 125.
We observe that, 4^{3 }= 64 < 91 < 125 = 5^{3}.
Hence 40^{3} = 64000 < 91125 < 125000 = 50^{3}. Hence n must lie between 40 and 50. Since the unit in the digits place is 5, therefore n must be 45.
ii) 166375
Solution:
Here unit in the digits place is 5. If n^{3} = 166375. Then, the unit in the digits place of n must be 5.
Let us split this as 166 and 375.
We observe that, 5^{3 }= 125 < 166 < 216 = 6^{3}.
Hence 50^{3} = 12500 < 166375 < 216000 = 60^{3}. Hence n must lie between 50 and 60. Since the unit in the digits place is 5, therefore n must be 55.
iii) 704969
Solution:
Here unit in the digits place is 9. If n^{3} = 704969. Then, the unit in the digits place of n must be 9.
Let us split this as 704 and 969.
We observe that, 8^{3 }= 512 < 704 < 729 = 9^{3}.
Hence 80^{3} = 512000 < 704969 < 729000 = 90^{3}. Hence n must lie between 80 and 90. Since the unit in the digits place is 9, therefore n must be 89.

Find the nearest integer to the cube root of each of the following.
i) 331776
Solution:
For easy simplification let us split 331776 as 331 and 776,
6^{3 }= 216 < 331 < 343 = 7^{3}
60^{3} = 216000 < 331776 < 343000 = 70^{3}.
Now it is closer to 70^{3 }than 60^{3}.
Let us go for more accurate, 69^{3 }= 328509 < 331776 < 34300 = 70^{3}.
Therefore , 331776 is closest to 69^{3}.
ii) 46656
Solution:
For easy simplification let us split 46656 as 46 and 656,
3^{3 }= 27 < 46 < 64 = 4^{3}
30^{3} = 27000 < 46656 < 64000 = 40^{3}.
Now it is closer to 40^{3 }than 30^{3}.
The number closer to 40 than 30 are 36, 37, 38, 39. Let us go through one by one.
36^{3 }= 46656, satisfies the condition.
iii. 373248
Solution:
For easy simplification let us split 372248 as 373 and 248,
7^{3 }= 343 < 373 < 512 = 8^{3}
70^{3} = 343000 < 373248 < 512000 = 80^{3}.
Now, it is closer to 70^{3 }than 80^{3}.
The numbers which are closer to 70 than 80 are : 71, 72, 73, 74, 75
Let us go for more accurate, 71^{3 }= 357911 < 373248 < 373248 = 72^{3}.
Therefore, 373248 is closest to 72^{3}
Squares, Square roots, Cubes and Cube roots
It’s been more than a decade since my last Math test, but I still have nightmares about it 🙂
The Curious Incident of the Dog in the Nighttime – the must book for you to read. I’ve always been fascinated by people who are good with numbers.
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