Download Cube nuMbeRs oR PeRfeCt Cubes

4
Cubes and Cube Roots
In earlier classes, you have learnt that a cube is a solid figure which has all its three sides equal
(i.e. length, breadth and height).
Volume of a cube = (side)3 (i.e. cube of side).
If the side of the cube is x units, then
volume of cube = (x × x × x) cubic units = x3 cubic units.
Cube of a number is the product of the number by itself three times.
73 = 7 × 7 × 7.
For example, 53 = 5 × 5 × 5,
In the previous chapter, you have learnt about squares and square roots, in this chapter we shall extend
our knowledge to cubes and cube roots.
Cube nuMbeRs oR PeRfeCt Cubes
Consider the following table:
Table 1
Natural number
1
2
3
4
5
6
7
8
9
10
Cube
1 =1×1×1=1
23 = 2 × 2 × 2 = 8
33 = 3 × 3 × 3 = 27
43 = 4 × 4 × 4 = 64
53 = 5 × 5 × 5 = 125
63 = 6 × 6 × 6 = 216
73 = 7 × 7 × 7 = 343
83 = 8 × 8 × 8 = 512
93 = 9 × 9 × 9 = 729
103 = 10 × 10 × 10 = 1000
3
You can see that 1, 8, 27, 64, 125, ….. are the natural numbers which are the cubes of natural numbers.
Such numbers are called cube numbers or perfect cubes.
Cubes of natural numbers are called cube numbers or perfect cubes.
In general, if a natural number m can be expressed as n3, where n is also a natural number, then m is
a cube number or perfect cube.
From table 1, we can write all the cube numbers between 1 and 1000. We find that the remaining
natural numbers are not perfect cubes.
Hence, we can say that:
All the natural numbers are not perfect cubes.
Cubes and Cube Roots
67
How do we know, whether a given natural number is a perfect cube?
Observe the following:
(i)Let us consider a cube number 64.
Expressing it into prime factors, we have
64 = 2 # 2 # 2 × 2 # 2 # 2 .
1 44 2 44 3
1 44 2 44 3
Observation: 64 can be expressed as the product of triplets of
equal prime factors.
(ii)Let us consider another cube number 729.
Expressing it into prime factors, we have
729 = 3 # 3 # 3 × 3 # 3 # 3
1 44 2 44 3
1 44 2 44 3
Observation: 729 can also be expressed as the product of triplets of
equal prime factors.
(iii)Now consider a non-cube number 192.
Expressing it into prime factors, we have
192 = 2 # 2 # 2 × 2 # 2 # 2 × 3
1 44 2 44 3
1 44 2 44 3
Observation: 192 cannot be expressed as the product of triplets of
equal prime factors.
Hence, we conclude that:
A perfect cube can always be expressed as the product of triplets of
equal prime factors.
Example 1. Is 243 a perfect cube?
Solution. Given number is 243
Expressing it into prime factors, we have
243 = 3 # 3 # 3 × 3 × 3
1 44 2 44 3
Since 3 × 3 left after grouping 3’s in triplets.
∴ 243 cannot be expressed as the product of triplets of equal prime factors.
Hence, 243 is not a perfect cube.
Example 2. Is 2744 a perfect cube?
Solution. Given number is 2744
Expressing it into prime factors, we have
2744 = 2 # 2 # 2 × 7 # 7 # 7
1 44 2 44 3
1 44 2 44 3
Since 2744 can be expressed as the product of triplets
of equal prime factors.
Hence, 2744 is a perfect cube.
Prime factorization
2
2
2
2
2
2
64
32
16
8
4
2
1
Prime factorization
3 729
3 243
3 81
3 27
3
9
3
3
1
Prime factorization
2 192
2 96
2 48
2 24
2 12
2
6
3
3
1
Prime factorization
3 243
3 81
3 27
3
9
3
3
1
Prime factorization
2
2
2
7
7
7
2744
1372
686
343
49
7
1
Learning Mathematics–VIII
68
Example 3. Is 675 a perfect cube? If not, find the smallest natural number by which 675 must be
multiplied so that product is a perfect cube.
Prime factorization
Solution. Given number is 675
Expressing it into prime factors, we have
675 = 5 × 5 × 3 # 3 # 3
1 44 2 44 3
Since 5 × 5 is left after grouping in triplets,
∴ 675 is not a perfect cube.
To make it perfect cube, 5 should also occur in a group of three, so we
need one more 5. In that case,
675 × 5 = 5 # 5 # 5 × 3 # 3 # 3 = 3375, which is a perfect cube.
1 44 2 44 3
5 675
5 135
3 27
3
9
3
3
1
1 44 2 44 3
Hence, the smallest number by which 675 must be multiplied so that product is a perfect cube is 5.
Example 4. Is 832 a perfect cube? If not, then by which smallest number should 832 be divided
so that the quotient is a perfect cube?
Solution. Given number is 832
Prime factorization
Expressing it into prime factors, we have
832 = 2 # 2 # 2 × 2 # 2 # 2 ×13.
1 44 2 44 3
1 44 2 44 3
1 44 2 44 3
1 44 2 44 3
Since 13 is left after grouping in triplets,
∴ 832 is not a perfect cube.
To make it perfect cube, we should divide the given number by 13,
then the prime factorization of the quotient will not contain 13.
In that case
832 ÷ 13 = 2 # 2 # 2 × 2 # 2 # 2 = 64, which is a perfect cube.
2
2
2
2
2
2
13
832
416
208
104
52
26
13
1
Hence, the smallest number by which 832 must be divided so that quotient is a perfect cube is 13.
Properties of cubes of natural numbers
Property 1. Consider the following table:
Table 2
Natural number
1
2
3
4
5
6
7
8
9
10
Cube
13 = 1
23 = 8
33 = 27
43 = 64
53 = 125
63 = 216
73 = 343
83 = 512
93 = 729
103 = 1000
Natural number
11
12
13
14
15
16
17
18
19
20
Cube
11 = 1331
123 = 1728
133 = 2197
143 = 2744
153 = 3375
163 = 4096
173 = 4913
183 = 5832
193 = 6859
203 = 8000
3
From the above table, we can observe that
13 = 1, 33 = 27, 53 = 125, 73 = 343, ….., 193 = 6859,…..…
i.e. cubes of odd numbers are odd.
Cubes and Cube Roots
69
and 23 = 8, 43 = 64, 63 = 216, 83 = 512, ……, 203 = 8000, ......
i.e. cubes of even numbers are even.
Hence, we can say that:
Cubes of odd numbers are odd and cubes of even numbers are even.
Property 2.
Study the table 2 and observe the unit place in a number and its cube. You will see that:
(i)If a number has 1, 4, 5, 6 or 9 in the unit place, then its cube also end with same digits i.e. 1,
4, 5, 6 or 9.
(ii)If a number has 2 in the unit place, then its cube ends in 8.
(iii)If a number has 8 in the unit place, then its cube ends in 2.
(iv)If a number has 3 in the unit place, then its cube ends in 7.
(v)If a number has 7 in the unit place, then its cube ends in 3.
(vi)If a number has 0 in the unit place, then its cube ends in 0.
Infact, it has three zeros in the end.
Cubes of negative integers
The cube of a negative integer is always negative.
For example:
(i)(– 2)3 = (– 2) × (– 2) × (– 2) = – 8
(ii)(– 5)3 = (– 5) × (– 5) × (– 5) = – 125
Cubes of rational numbers
p
q
If
(q ≠ 0) is a rational number, then
p 3
p
c m =
q
q
∴
p
c m
q
3
p
=
p
q
×
×
p
q
=
p#p#p
q#q#q
=
p3
q3
3
q3
For example:
3
3
(i) c 2 m = 23 = 2 # 2 # 2 = 8 .
3
(ii) c − 4 m =
3
5
3#3#3
3
(− 4)
5
3
3
=
27
(− 4) # (− 4) # (− 4)
5#5#5
= − 64 .
125
Exercise 4.1
1. Which of the following numbers are not perfect cubes? Give reasons in support of your
answer:
(i) 648
(ii) 729
(iii) 8640
(iv) 1000
(v) 3000
2. Show that each of the following numbers is a perfect cube. Also find the number whose cube
is the given number:
(i) 1728
(ii) 5832
(iii) 13824
(iv) 59319
(v) 91125
3. Find the smallest number by which each of the following numbers must be multiplied to obtain
a perfect cube:
(i) 243
(ii) 3072
(iii) 11979
(iv) 19652