If a number is multiplied by itself three times, then the product is called the cube of that number.
To denote the cube of a number, 3 is written a little up to the right of the numbers.
Thus, 3 is the power of x and is read as “x cubed”.
When a number is multiplied three times by itself, the product obtained is called the cube of a number.
For a given numberm, we define, cube of m = m × m × m, denoted by m³.
For example:
(i) 2³ = (2 × 2 × 2) = 8.
Thus, cube of 2 is 8.
(ii) 3³ = (3 × 3 × 3) = 27.
Thus, cube of 3 is 27.
(iii) 4 × 4 × 4 = 64, here 64 is the cube of 4.
(iv) 5 × 5 × 5 = 125, here 125 is the cube of 5.
A natural number (n) is said to be a perfect cube if (n = m³) it is the cube of some natural number.
For example:
1³ =1,
2³ = 8,
3³ =27,
4³ =64,
5³ =125, etc.
Thus 1, 8, 27, 64, 125, etc. are perfect cubes.
A given natural number is a perfect cube if it can be expressed as the product of triplets of equal factors.
The cube of a negative integer is always negative.
For example:
(-1)³ = (-1) × (-1) × (-1) = -1,
(-2)³ = (-2) × (-2) × (-2) = -8
(-3)³ = (-3) × (-3) × (-3) = -27, etc.
We have, (a/b) ³ = a/b × a/b × a/b = (a × a × a)/(b × b × b) = a³/b³
Hence, (a/b) ³ = a³/ b³
For example:
(i) (3/5) ³ = 3³/5 ³ = (3 × 3 × 3)/(5 × 5 × 5) = 27/125
(ii) (-2/3) ³ = (-2) ³/ 3³ = {(-2) × (-2) × (-2)}/(3 × 3 × 3) = -8/27
(i) The cube of every even natural number is even.
(ii) The cube of every odd natural number is odd.
1. Show that 189 is not a perfect cube.
Solution:
Resolving 189 into prime factors, we get:
189 = 3 × 3 × 3 × 7
Making triplets, we find that one triplet is formed and we are left with one more factor.
Thus, 189 cannot be expressed as a product of triplets.
Hence, 189 is not a perfect cube.
2. Show that 216 is a perfect cube. Find the number whose cube is 216.
Solution:
Resolving 216 into prime factors, we get:
216 = 2 × 2 × 2 × 3 × 3 × 3
= (2 × 3) × (2 × 3) × (2 × 3)
= (6 × 6 × 6)
= 6³
Thus, 216 is a perfect cube.
And, 6 is the number whose cube is 216.
3. What is the smallest number by which 3087 may be multiplied so that the product is a perfect cube?
Solution:
Writing 3087 as a product of prime factors, we have:
3087 = 3 × 3 × 7 × 7 × 7
Hence, to make it a perfect cube, it must be multiplied by 3.
4. What is the smallest number by which 392 may be divided so that the quotient is a perfect cube?
Solution:
Writing 392 as a product of prime factors, we have:
392 = 2 × 2 × 2 × 7 × 7
Clearly, to make it a perfect cube, it must be divided by (7 × 7), i.e., 49.
5. Find the cube of each of the following :
(i) (-70 ) (ii) 1²/₃ (iii) 2.5 (iv) 0.06
Solution:
(i) (-7)³
= (-7) × (-7) × (-7)
= -343
(ii) (1²/₃)³
= (5/3) ³
= 5³/3³
= (5 × 5 × 5)/(3 × 3 × 3)
= 125/27
(iii) (2.5)³
= (25/10)³
= (5/2)³
= 5³/3³
= (5× 5 × 5)/(3× 3× 3)
= 125/27
(iv) (0.06) ³
= (6/100)³
= (3/50)³ = 3³/(50)³
= (3 × 3 × 3)/(50 × 50 × 50)
= 27/125000
● Cube and Cube Roots
To Find if the Given Number is a Perfect Cube
Method for Finding the Cube of a Two-Digit Number
● Cube and Cube Roots - Worksheets
Worksheet on Cube and Cube Root
8th Grade Math Practice
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