## 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.