Maths Chapter-07: Cubes and Cube Roots
Exercise 7.1
Q1:
Mention the numbers that are not perfect cubes.
(A) 216
(B) 128
(C) 1000
(D) 100
(E) 46656
Solution:
(A) 216
Prime factors of 216: 2x2x2x3x3x3
Here all the factors are in the groups of 3’s
Therefore, 216 is said to be a perfect cube number.
| 02 | 0216 |
| 02 | 0108 |
| 02 | 054 |
| 03 | 027 |
| 03 | 09 |
| 03 | 03 |
| 01 |
(B) 128
The prime factor of 128 = 2x2x2x2x2x2x2
Here one factor 2 does not appear in groups of 3
Hence, 128 is not a perfect cube.
| 02 | 0128 |
| 02 | 064 |
| 02 | 032 |
| 02 | 016 |
| 02 | 08 |
| 02 | 04 |
| 02 | 02 |
| 01 |
(C) 1000
The prime factors of 1000 = 2x2x2x 5x5x5
Here all the factors are in groups of 3
Hence, 1000 is said to be a perfect cube.
| 02 | 01000 |
| 02 | 0500 |
| 02 | 0250 |
| 05 | 0125 |
| 05 | 025 |
| 05 | 05 |
| 01 |
(D) 100
The prime factors of 100 is 2×2 x 5×5
Here all the factors do not appear in groups of 3.
Hence, 100 is not a perfect cube.
| 02 | 0100 |
| 02 | 050 |
| 05 | 025 |
| 05 | 05 |
| 01 |
(E) 46656
The prime factors of 46656 = 2x2x2x2x2x2x 3 x3x3x3x3x 3
Here all the factors are in groups of 3
Hence, 46656 is said to be a perfect cube.
| 02 | 046656 |
| 02 | 023328 |
| 02 | 011664 |
| 02 | 05832 |
| 02 | 02916 |
| 02 | 01458 |
| 03 | 0729 |
| 03 | 0243 |
| 03 | 081 |
| 03 | 027 |
| 03 | 09 |
| 03 | 03 |
| 01 |
Q2 :
Find the smallest number when multiplied to obtain a perfect cube:
(A) 243
(B) 256
(C) 72
(D) 675
(E) 100
Solution:
(A) 243
The prime factors of 243 = 3x3x3x3x 3
Here 3 does not appear in groups of 3
Hence, For 243 to be a perfect cube it should be multiplied by 3.
| 03 | 0243 |
| 03 | 081 |
| 03 | 027 |
| 03 | 09 |
| 03 | 03 |
| 01 |
(B) 256
The prime factors of 256 is 2x2x2x2x2x 2 x2 x 2
Here one factor of 2 is required for it to make groups of 3.
Hence, for 256 to be a perfect cube it should be multiplied by 2.
| 02 | 0256 |
| 02 | 0128 |
| 02 | 064 |
| 02 | 032 |
| 02 | 016 |
| 02 | 08 |
| 02 | 04 |
| 02 | 02 |
| 01 |
(C) 72
The prime factors for 72 = 2 x2x2x 3x 3
Here the factor 3 does not appear in groups of 3
Hence, For 72 to be a perfect cube it should be multiplied by 3.
(D) 675
The prime factors for 675 = 3x3x3x 5×5
Here the factor 5 does not appear in groups of 3
Hence, for 675 to be a perfect cube it should be multiplied by 5.
| 03 | 0675 |
| 03 | 0225 |
| 03 | 075 |
| 05 | 025 |
| 05 | 05 |
| 01 |
(E) 100
The prime factors for 100 = 2x2x5x5
Here both the factors 2 and 5 are not in groups of 3
Hence, for 100 to be a perfect cube it should be multiplied by 2 and 5. ( i.e. 2 x 5 =10 )
| 02 | 0100 |
| 02 | 050 |
| 05 | 025 |
| 05 | 05 |
| 01 |
Q3:
Find the smallest number by which when divided obtain a perfect cube.
(A) 81
(B) 128
(C) 135
(D) 192
(E) 704
Solution:
(A) 81
The prime factors for 81 = 3 x 3 x 3 x 3
Here, there is one factor of 3 which extra from the group of 3
Hence, for 81 to be a perfect cube it should be divided by 3.
| 03 | 081 |
| 03 | 027 |
| 03 | 09 |
| 03 | 03 |
| 01 |
(B) 128
The prime factors of 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here there is one factor of 2 which in not in the group of 3
Hence, for 128 to be a perfect cube then it should be divided by 2.
| 02 | 0128 |
| 02 | 064 |
| 02 | 032 |
| 02 | 016 |
| 02 | 08 |
| 02 | 04 |
| 02 | 02 |
| 01 |
(C) 135
The prime factors of 135 = 3 x 3 x 3 x 5
Here there is one factor of 5 which is not appearing with its group of 3.
Hence, for 135 to be a perfect cube it should be divided by 5.
| 03 | 0135 |
| 03 | 045 |
| 03 | 15 |
| 05 | 05 |
| 01 |
(D)192
The prime factors for 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3
Here there is one factor of 3 which does not appearing with its group of 3.
Hence for 192 to be a perfect cube then it should be divided by 3.
| 02 | 0192 |
| 02 | 096 |
| 02 | 048 |
| 02 | 024 |
| 02 | 012 |
| 02 | 06 |
| 03 | 03 |
| 01 |
(E) 704
The prime factor for 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11
Here there is one factor of 11 which is not appearing with its group of 3.
Hence for 704 to be a perfect cube it should be divided by 11.
| 02 | 0704 |
| 02 | 0352 |
| 02 | 0176 |
| 02 | 088 |
| 02 | 044 |
| 02 | 022 |
| 02 | 011 |
| 01 |
Q4:
Reuben makes a cuboid of clay of sides 5 cm , 2 cm , 5 cm. If Reuben wants to form a cube how many such cuboids will be needed?
Solution:
The numbers given: 5 x 2 x 5
Since the factors of 2 and 4 are both not in groups of 3.
Then, the number should be multiplied by 2 x 2 x 5 = 20 for it to be made a perfect cube.
Hence Reuben needs 20 cuboids.
Exercise 7.2
Q1 :
By the method of prime factorization find the cube root for the following.
(A) 64
(B) 512
(C) 10648
(D) 27000
(E) 15625
(F) 13824
(G) 110592
(H) 46656
(I) 175616
(J) 91125
Solution:
(A) 64
\(\sqrt[3]{64}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\\\sqrt[3]{64}=2\times 2\)
= 4
| 02 | 064 |
| 02 | 032 |
| 02 | 016 |
| 02 | 08 |
| 02 | 04 |
| 02 | 02 |
| 01 |
(B) 512
\(\sqrt[3]{512}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2}\)
= 2 x 2 x 2
= 8
| 02 | 0512 |
| 02 | 0256 |
| 02 | 0128 |
| 02 | 064 |
| 02 | 032 |
| 02 | 016 |
| 02 | 08 |
| 02 | 04 |
| 02 | 02 |
| 01 |
(C) 10648
\sqrt[3]{10648}=\sqrt[3]{2\times 2\times 2\times 11\times 11\times 11}
= 2 x 11
=22
| 02 | 010648 |
| 02 | 05324 |
| 02 | 02662 |
| 011 | 01331 |
| 011 | 0121 |
| 011 | 011 |
| 01 |
(D) 27000
\(\sqrt[3]{27000}=\sqrt[3]{2\times 2\times 2\times 3\times 3\times 3\times 5\times 5\times 5}\)
=>2 x 3 x 5
=>30
| 02 | 027000 |
| 02 | 013500 |
| 02 | 06750 |
| 03 | 03375 |
| 03 | 01125 |
| 03 | 0375 |
| 05 | 0125 |
| 05 | 025 |
| 05 | 05 |
| 01 |
(E) 15625
\(\sqrt[3]{15625}=\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5}\)
=> 5 x 5
=> 25
| 05 | 015625 |
| 05 | 03125 |
| 05 | 0625 |
| 05 | 0125 |
| 05 | 025 |
| 05 | 05 |
| 01 |
(F) 13824
\(\sqrt[3]{13824}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3}\)
=> 2 x 2 x 2 x 3
=> 24
| 02 | 13824 |
| 02 | 06912 |
| 02 | 03456 |
| 02 | 01728 |
| 02 | 0864 |
| 02 | 0432 |
| 02 | 0216 |
| 02 | 0108 |
| 02 | 054 |
| 03 | 27 |
| 03 | 09 |
| 03 | 03 |
| 01 |
(G) 110592
\(\sqrt[3]{110592}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \times3\times 3\times 3}\)
=> 2 x 2 x 2 x 2 x 3
=> 48
| 02 | 0110592 |
| 02 | 055296 |
| 02 | 027648 |
| 02 | 013824 |
| 02 | 06912 |
| 02 | 03456 |
| 02 | 01728 |
| 02 | 0864 |
| 02 | 0432 |
| 02 | 0216 |
| 02 | 0108 |
| 02 | 054 |
| 03 | 027 |
| 03 | 09 |
| 03 | 03 |
| 01 |
(H) 46656
\(\sqrt[3]{46656}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3}\)
=> 2 x 2 x 2 x 3 x 3 x 3
=> 36
(I) 175616
\(\sqrt[3]{175616}=\sqrt[3]{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7\times 7\times 7}\)
=> 2 x 2 x 2 x 7
=> 56
| 02 | 0175616 |
| 02 | 087808 |
| 02 | 043904 |
| 02 | 021952 |
| 02 | 010976 |
| 02 | 05488 |
| 02 | 02744 |
| 02 | 01372 |
| 02 | 0686 |
| 07 | 0343 |
| 07 | 049 |
| 07 | 07 |
| 01 |
(J) 91125
\(\sqrt[3]{91125}=\sqrt[3]{3\times 3\times 3\times 3\times 3\times 3\times 5\times 5\times 5}\)
=> 3 x 3 x 5
=> 45
| 03 | 091125 |
| 03 | 010125 |
| 03 | 03375 |
| 03 | 01125 |
| 03 | 0375 |
| 05 | 0125 |
| 05 | 025 |
| 05 | 05 |
| 01 |
Q2:
State whether the following is true of false:
(A) Any off number of a cube is even.
(B) When a number end with two zeros, it is never a perfect cube.
(C) If the square of a given number ends with 5 then its cube will end with 25.
(D) There is no number that ends with 8 which is a perfect cube.
(E) The cube of a given two digit number will always be a three digit number.
(F) The cube of a two digit number will have either seven or more digits.
(G) The cube of single digit number may also be a single digit number.
Solution:
(A) The statement given is false.
Since, 13 = 1, 33 = 27, 53 = 125, . . . . . . . . . . are all odd.
(B) The given statement is true.
Since, a perfect cube ends with three zeroes.
Eg. 103 = 1000, 203 = 8000, 303 = 27,000 , . . . . . . . . . . . . so on.
(C) The given statement is false
Since, 52 = 25, 53 = 125 , 152 = 225, 153 = 3375 ( Did not end with 25)
(D) the given statement is false.
Since 123 = 1728 [the number ends with 8]
223 = 10648 [ the number ends with 8]
(E) The given statement is false
Since, 103 = 1000 [Four digit number]
And 113= 1331 [four digit number]
(F) The statement is False.
Since 993 = 970299 [Six digit number]
(G) the given statement is true
13 = 1 [single digit]
23 = 8 [single digit]
Q3 :
1331 is told to be a perfect cube. What are the factorization methods in which you can find its cube root? Similarly, find the cube roots for
(i)4913
(ii)12167
(iii)32768.
Solution:
We know that 103 = 1000 and possible cute of 113 = 1331
Since, the cube of units digit is 13 = 1
Then, cube root of 1331 is 11
(i) 4913
We know that 73 is 343
Next number that comes with 7 as the units place is 173 = 4913
Therefore the cube root of 4913 is 17
(ii) 12167
Since we know that 33 = 27
Here in cube, the ones digit is 7
Now the next number with 3 In the ones digit is 133 = 2197
And the next number with 3 in the ones digit is 233 = 12167
Hence the cube root of 12167 is 23
(iii) 32768
We know that 23 = 8
Here in the cube, the ones digit is 8
Now the next number with 2 in the ones digit is 123 = 1728
And the next number with 2 as the ones digit 223 = 10648
And the next number with 2 as the ones digit 323 = 32768
Hence the cube root of 32768 is 32