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We know that the word 'Cube' is used in geometry. A cube is a solid figure which has all its sides are equal.

If the side of a cube in the adjoining figure is βaβ units then its volume is given by
= a β a β a
= a3 cubic units.
Here a3 is called 'a cubed' or 'a raised to the power 3' or 'a to the power 3'.
Now, consider the number 1, 8, 27, 64, 125,.......
These are called perfect cubes or cube numbers.
Each of them is obtained when a number is multiplied by itself three times.
Examples :
1 β 1 β 1 = 13
2 β 2 β 2 = 23
3 β 3 β 3 = 33
5 β 5 β 5 = 53
The following are the cubes of numbers from 11 to 20.

From the above table we observe the following properties of cubes :
Property 1 :
For numbers with their unitβs digit as 1, their cubes also will have the unitβs digit as 1.
For example,
13 = 1
113 = 1331
213 = 9261
313 = 29791
Property 2 :
The cubes of the numbers with 1, 4, 5, 6, 9 and 0 as unit digits will have the same unit digits.
For example,
143 = 2744
153 = 3375
163 = 4096
203 = 8000
Property 3 :
The cube of numbers ending in unit digit 2 will have a unit digit 8 and the cube of the numbers ending in unit digit 8 will have a unit digit 2.
For example,
123 = 1728
183 = 5832
Property 4 :
The cube of the numbers with unit digits as 3 will have a unit digit 7 and the cube of numbers with unit digit 7 will have a unit digit 3.
For example,
133 = 2197
273 = 19683
Property 5 :
The cubes of even numbers are all even; and the cubes of odd numbers are all odd.
For example,
183 = 5832 (even)
273 = 19683 (odd)
Property 6 :
The sum of the cubes of first n natural numbers is equal to the square of their sum.
13 + 23 + 33 + ....... + n3 = (1 + 2 + 3 + ..... + n)2
For example,
13 + 23 + 33 + 43 = (1 + 2 + 3 + 4)2
1 + 8 + 27 + 64 = 102
100 = 100
Problem 1 :
Is 64 a perfect cube ?
Solution :
Using prime factorization, we can write 64 as given below.
64 = 2 β 2 β 2 β 2 β 2 β 2
64 = 23 β 23 = (2 β 2)3 = 43
Hence, 64 is a perfect cube.
Problem 2 :
Is 500 a perfect cube ?
Solution :
Using prime factorization, we can write 500 as given below.
500 = 2 β 2 β 5 β 5 β 5
There are three 5βs in the product but only two 2βs.
Hence, 500 is not a perfect cube.
Problem 1 :
The cube root of 0.000216 is
a) 0.6 b) 0.06 c) 0.006 d) none
Solution :
= β0.000216
= β(216/1000000)
= β(6 x 6 x 6) / (100 x 100 x 100)
= 6/100
= 0.06
So, option b is correct.
Problem 2 :
β(4 12/125) = ?
a) 1 2/5 b) 1 3/5 c) 1 4/5 d) 2 2/5
Solution :
= β(4 12/125)
Converting the mixed fraction as improper fraction, we get
= β(4 x 125 + 12)/125
= β(500 + 12)/125
= β512/125
= = β(8 x 8 x 8)/(5 x 5 x 5)
= 8/5
Converting improper fraction, we get
= 1 3/5
So, option b is correct.
Problem 3 :
β(β0.000064) = ?
a) 0.02 b) 0.2 c) 2 d) None
Solution :
= β(β0.000064)
First we will try to remove the square root.
= ββ(64/1000000)
= ββ(8 x 8)/(1000 x 1000)
When two same values are multiplied, we may take one value out of the square root.
= β8/1000
= β(2 x 2 x 2)/(10 x 10 x 10)
= 2/10
= 0.2
Problem 4 :
The largest four digit number which is perfect cube is
a) 8000 b) 9261 c) 9999 d) None
Solution :
Option a :
= β8000
= β2 x 2 x 2 x 10 x 10 x 10
= 2 x 10
= 20
Option b :
= β9261
= β3 x 3 x 3 x 7 x 7 x 7
= 3 x 7
= 21
Option c :
= β9999
= β3 x 3 x 1111
It is not a perfect cube.
So, option a is correct.
Problem 5 :
What is the smallest number by which 3600 be divided to make it a perfect cube.
a) 9 b) 50 c) 300 d) 450
Solution :
= β3600
= β2 x 2 x 2 x (2 x 5 x 5 x 3 x 3)
(2 x 5 x 5 x 3 x 3) is the number to be divided to make it as perfect cube.
So, 450 is the number to be removed to make it as perfect cube.
Problem 6 :
A rectangular solid has a length three times the width and a height twice its width. If the volume of the rectangle solid is 384 ππ3, determine the dimensions of the rectangular solid.
Solution :
Let x be the width
length = 3x
height = 2x
Volume of rectangular prism = length x width x height
x (3x)(2x) = 384 ππ3
6x3 = 384
x3 = 384/6
x3 = 64
x = β64
x = β(4 x 4 x 4)
x = 4
length = 3x ==> 12 cm
width = 4 cm
height = 2x ==> 8 cm
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