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

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

The document RD Sharma Solutions for Class 8 Math Chapter 4 - Cubes and Cube Roots (Part-1) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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#### 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 1 1 Odd 2 8 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 3 27 Odd (Not an even number) 4 64 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 5 125 Odd (Not an even number) 6 216 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 7 343 Odd (Not an even number) 8 512 Even (Last digit is even, i.e., 0, 2, 4, 6, 8) 9 729 Odd (Not an even number) 10 1000 Even (Last digit is even, i.e., 0, 2, 4, 6, 8)

#### Question 3: Observe the following pattern:                13 = 1        13 + 23 = (1 + 2)213 + 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.

#### 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.