RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

Class 8 : RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

Question 1: Find the  cubes of the following numbers:
(i) 7
(ii) 12
(iii) 16
(iv) 21
(v) 40
(vi) 55
(vii) 100
(viii) 302
(ix) 301

Answer 1: Cube of a number is given by the number raised to the power three.
(i) Cube of 7 = 73 = 7×7×7 =343Cube of 7 = 73 = 7×7×7 =343
(ii)  Cube of 12 = 123 = 12×12×12 = 1728Cube of 12 = 123 = 12×12×12 = 1728
(iii) Cube of 16 =163 = 16×16×16 = 4096Cube of 16 =163 = 16×16×16 = 4096
(iv) Cube of 21 = 213 = 21×21×21 = 9261Cube of 21 = 213 = 21×21×21 = 9261
(v) Cube of 40 = 403 = 40×40×40 = 64000Cube of 40 = 403 = 40×40×40 = 64000
(vi) Cube of 55 = 553 = 55×55×55 = 166375Cube of 55 = 553 = 55×55×55 = 166375
(vii) Cube of 100 = 1003 = 100×100×100 = 1000000Cube of 100 = 1003 = 100×100×100 = 1000000
(viii) Cube of 302 = 3023 = 302×302×302 = 27543608Cube of 302 = 3023 = 302×302×302 = 27543608
(ix) Cube of 301 = 3013 = 301×301×301 =27270901Cube of 301 = 3013 = 301×301×301 =27270901 

Question 2: Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.

Answer 2: The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
(i) From the above table, it is evident that cubes of all odd natural numbers are odd.
(ii) From the above table, it is evident that cubes of all even natural numbers are even. 

Number 
Cube 
Classification 
11
Odd 
28Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 
327Odd (Not an even number) 
464Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 
5125Odd (Not an even number) 
6216Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 
7343Odd (Not an even number) 
8512Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 
9729Odd (Not an even number) 
101000Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 

Question 3: Observe the following pattern:
                13 = 1
        13 + 23 = (1 + 2)2
13 + 23 + 33 = (1 + 2 + 3)2 

Write the next three rows and calculate the value of 13 + 2+ 33 + ... + 9+ 103 by the above pattern. 

Answer 3: Extend the pattern as follows: 


 
                                  13=1
                            13+23=(1+2)2                     13+23+33=(1+2+3)2              13+23+33+43=(1+2+3+4)2       13+23+33+43+53=(1+2+3+4+5)213+23+33+43+53+63=(1+2+3+4+5+6)2

Now, from the above pattern, the required value is given by: 

13+23+33+43+53+63+73+83+93+103=(1+2+3+4+5+6+7+8+9+10)2=552=302513+23+33+43+53+63+73+83+93+103=1+2+3+4+5+6+7+8+9+102=552=30Thus, the required value is 3025. 

Question 4: Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
'The cube of a natural number which is a multiple of 3 is a multiple of 27'

Answer 4: Five natural numbers, which are multiples of 3, are 3, 6, 9, 12 and 15.
Cubes of these five numbers are: 

33 = 3×3×3 = 2763 = 6×6×6 = 21693 = 9×9×9 = 729123 = 12×12×12 = 1728153 = 15×15×15 = 3375

Now, let us write the cubes as a multiple of 27. We have: 

27 = 27 × 1216 = 27 ×8729 = 27 × 271728 = 27 × 643375 = 27 × 125It is evident that the cubes of the above multiples of 3 could be written as multiples of 27. Thus, it is verified that the cube of a natural number, which is a multiple of 3, is a multiple of 27. 

Question 5: Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, ...) and verify the following:
'The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1'.

Answer 5: Five natural numbers of the form (3n + 1) could be written by choosing n=1,2,3... etc.n=1,2,3... etc.
Let five such numbers be 4,7,10,13, and 16.4,7,10,13, and 16.

The cubes of these five numbers are:
43=64, 73=343, 103=1000, 133=2197 and 163=409643=64, 73=343, 103=1000, 133=2197 and 163=4096
The cubes of the numbers 4,7,10,13, and 164,7,10,13, and 16  could expressed as:
    64=3×21+164=3×21+1, which is of the form (3n + 1) for = 21
  343=3×114+1343=3×114+1, which is of the form (3n + 1) for = 114
1000=3×333+1,1000=3×333+1, which is of the form (3n + 1) for = 333
2197=3×732+1, 2197=3×732+1, which is of the form (3n + 1) for = 732
4096=3×1365+1,4096=3×1365+1, which is of the form (3n + 1) for = 1365
The cubes of the numbers 4,7,10,13, and 164,7,10,13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified. 

Question 6: Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11, ...) and verify the following:
'The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2'.

Answer 6: Five natural numbers of the form (3n + 2) could be written by choosing n=1,2,3... etc.n=1,2,3... etc.
Let five such numbers be 5,8,11,14, and 17.5,8,11,14, and 17.
The cubes of these five numbers are: 
53=125, 83=512, 113=1331, 143=2744, and 173=4913.53=125, 83=512, 113=1331, 143=2744, and 173=4913.
The cubes of the numbers 5,8,11,14 and 175,8,11,14 and 17 could expressed as:
    125=3×41+2125=3×41+2, which is of the form (3n + 2) for