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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”.



Cube of a number:

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.


Perfect cube:

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.


Cubes of negative integer:

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.


Cube of a rational number:

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


Properties of cubes of numbers:

(i) The cube of every even natural number is even.


(ii) The cube of every odd natural number is odd.



Solved example to find perfect cubes step by step;

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

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