Cubes

The cube of a number is the number raised to the power 3. Thus,
cube of 2 = \( { 2 }^{ 3 } \) = 2 x 2 x 2 = 8,
cube of 5 = \( { 2 }^{ 3 } \) = 5 x 5 x 5 = 125.

Perfect Cube


We know that \( { 2 }^{ 3 } \) = 8, \( { 3 }^{ 3 } \) = 27, \( { 26}^{ 3 } \) = 216, \( { 7 }^{ 3 } \) = 343, \( { 10 }^{ 3 } \)= 1000.
The numbers 8, 27, 216, 343, 1000, … are called perfect cubes. A natural number is said to be a perfect cube, if it is the cube of some natural number, i.e.,
A natural number n is a perfect cube if there exists a natural number m such that  m X m X m = n.

Procedure :

Step I- Obtain the natural number.

Step II- Express the given natural number as a product of prime factors.

Step III- Group the factors in triples in such a way that all the three factors in each triple are equal.

Step IV- If no factor is left over in grouping in step III, then the number is a perfect cube, otherwise not.

Illustrative Examples :

Example 1: Show that 729 is a perfect cube.
Solution. Resolving 729 into prime factors, we have
                     729 =3 x 3 x 3 x 3 x 3 x 3
Here, we find that the prime factor 3 of the given number can be grouped into triplets and no factor is left out. Hence, 729 is a perfect cube.
Also, 729 is the cube of 3 x 3, i.e., 729 = \( { (9) }^{ 3 } \)•

Example 2: What is the smallest number by which 1323 may be multiplied so that the product is a perfect cube?
Solution. Resolving 1323 into prime factors, we have
                    1323 = 3 x 3 x 3 x 7 x 7
Since one more 7 is required to make a triplet of 7, the smallest number by which 1323 should be multiplied to make it a perfect cube is 7.

Example 3: What is the smallest number by which 1375 should be divided so that the quotient may be a perfect cube?
Solution. Resolving 1375 into prime factors, we have
                    1375 = 5 x 5 x 5 x 11
The factor 5 makes a triplet, and 11 is left out. So, clearly 1375 should be divided by 11 to make it a perfect cube.

Properties of Cubes of Numbers

1. Cubes of all odd natural numbers are odd. Thus  \( { 3 }^{ 3 } \) = 27, \( { 5 }^{ 3 } \) = 125, \( { 7 }^{ 3 } \) = 343, \( { 9 }^{ 3 } \) = 729, etc.

 

2. Cubes of all even natural numbers are even. Thus \( { 2 }^{ 3 } \) = 8, \( { 4 }^{ 3 } \) = 64, \( { 6 }^{ 3 } \) = 216, \( { 8 }^{ 3 } \) = 512, etc.

3. The cube of a negative integer is always negative
e.g.,  \( { (-1) }^{ 3 } \) = (—1)x (—1) x (—1) = (1) x (—1) = —1.
         \( { (-2) }^{ 3 } \) = -2 x -2 x -2 =(-2 X -2) X -2 = 4 X —2 = -8.

 

4. For any rational number \( \frac { a } { b } \), we have \( { (\frac { a } { b } ) }^{ 3 }=(\frac { { a }^{ 3 } } { { b }^{ 3 } } ) \).
Thus, \( { (\frac { 2 }{ 3 } ) }^{ 3 }=(\frac { { 2 }^{ 3 } }{ { 3 }^{ 3 } } ) \) = \( \frac { 8 }{ 27 } \) .

  \( { (\frac { -4 }{ 5 } ) }^{ 3 }=(\frac { { (-4) }^{ 3 } }{ { 5 }^{ 3 } } ) \) = \( \frac { -64 }{ 125 } \) .

5. The sum of the cubes of first n natural numbers is equal to the square of their sum. That is,

\( { 1 }^{ 3 } \) + \( { 2 }^{ 3 } \) + \( { 3 }^{ 3 } \) … + \( { n }^{ 3 } \) = \( { (1 + 2 + 3+ … + n) }^{ 2 } \)

6. Cubes of the numbers ending in digits 1, 4, 5, 6 and 9 are the numbers ending in the same digit. Cubes of numbers ending in digit 2 ends in digit 8 and the cube of numbers ending in digit 8 ends in digit 2. The cubes of the numbers ending in digits 3 and 7 ends in digit 7 and 3 respectively.

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