RBSE Solutions Class 8 Maths Chapter 2 Cube and Cube Roots More Ques

Question 1. Fill in the blanks
Answer:

The table shows that the number of unit cubes making a large cube is the cube of the number of unit cubes on one side. This is because a cube's volume is found by multiplying its side length by itself three times.
In simple words: The number of small cubes needed to build a bigger cube is found by cubing the number of small cubes along one edge.

🎯 Exam Tip: Remember that a cube has equal length, width, and height, so the total number of unit cubes is always the side length multiplied by itself three times.

Question 2. Find the cube and fill in the blanks
Answer:

To find the cube of a number, you multiply the number by itself three times. This operation is fundamental in understanding three-dimensional space.
In simple words: To "cube" a number means to multiply it by itself three times. For example, the cube of 2 is \( 2 \times 2 \times 2 = 8 \).

🎯 Exam Tip: Knowing the cubes of numbers from 1 to 10 by heart can save a lot of time in calculations.

Question 3. We know that \( 2^2 = 4 \) where \( 4 = 2 \times 2 \) or \( 4 = 2 + ? \) Similarly, \( 2^3 = 8 \) where \( 8 = 2 \times 2 \times 2 \). Does this equal to \( (2 + 2 + 2) \)?
Answer: We know that \( 2^3 \) means \( 2 \times 2 \times 2 \). When we multiply these numbers, we get \( 8 \). However, if we add 2 three times, \( 2 + 2 + 2 \), the result is \( 6 \). Since \( 8 \ne 6 \), we can see that \( 2^3 \) is not equal to \( 2 + 2 + 2 \). This shows that cubing a number is a different operation from repeated addition.
In simple words: Cubing a number (like \( 2^3 \)) means multiplying it by itself three times, which gives 8. Adding the number three times (like \( 2+2+2 \)) gives 6. So, \( 2^3 \) is not the same as \( 2+2+2 \).

🎯 Exam Tip: Always remember the difference between multiplication and addition. Squaring or cubing is repeated multiplication, not repeated addition.

Question 4. Find the unit digit of cubes given below
(i) 1331
(ii) 4444
(iii) 159
(iv) 1005
Answer:
(i) For the number 1331, the unit digit is 1. The cube of any number ending in 1 will also have 1 as its unit digit. So, the unit digit of \( 1331^3 \) is 1.
(ii) For the number 4444, the unit digit is 4. The cube of any number ending in 4 will also have 4 as its unit digit. So, the unit digit of \( 4444^3 \) is 4.
(iii) For the number 159, the unit digit is 9. The cube of any number ending in 9 will also have 9 as its unit digit. So, the unit digit of \( 159^3 \) is 9.
(iv) For the number 1005, the unit digit is 5. The cube of any number ending in 5 will also have 5 as its unit digit. So, the unit digit of \( 1005^3 \) is 5.
In simple words: To find the unit digit of a cube, just look at the unit digit of the original number and find its cube. The unit digit of that cube is your answer. Numbers ending in 0, 1, 4, 5, 6, 9 keep their unit digit when cubed.

🎯 Exam Tip: The unit digit of a number's cube is determined solely by the unit digit of the number itself. Memorize the pattern for 2 (ends in 8), 3 (ends in 7), 7 (ends in 3), 8 (ends in 2), and for others, the unit digit remains the same.

Question 5. The Cube of 46 will be even or odd.
Answer: The number 46 is an even number. When an even number is cubed, the result will always be an even number. This is because multiplying even numbers together always produces an even number (even x even = even).
In simple words: Because 46 is an even number, its cube will also be an even number. Even numbers always make even cubes.

🎯 Exam Tip: A simple rule to remember: the cube of an even number is always even, and the cube of an odd number is always odd.

Question 6. Observe the following pattern of sums of odd numbers.
\( 1 = 1 = 1^3 \)
\( 3 + 5 = 8 = 2^3 \)
\( 7 + 9 + 11 = 27 = 3^3 \)
\( 13 + 15 + 17 + 19 = 64 = 4^3 \)
\( 21 + 23 + 25 + 27 + 29 = 125 = 5^3 \)
Think on this pattern tell : How many consecutive odd numbers will be needed to get the sum of \( 10^3 \)?
Answer: Looking at the pattern, we can see that to get the cube of a number \( n \), we need to add \( n \) consecutive odd numbers. For example, for \( 1^3 \), we use 1 odd number; for \( 2^3 \), we use 2 odd numbers; for \( 3^3 \), we use 3 odd numbers, and so on. Following this pattern, to get the sum of \( 10^3 \), we will need 10 consecutive odd numbers. This pattern helps visualize how cubes are built from odd number sequences.
In simple words: To get the cube of any number, you need to add that many odd numbers in a row. So, for \( 10^3 \), you will need 10 odd numbers.

🎯 Exam Tip: The first odd number in the sum for \( n^3 \) is given by the formula \( n(n-1) + 1 \). This helps you find the starting point for any number's cube in this pattern.

Question 7. According to the above pattern, find the below in form of addition of odd numbers.
(i) \( 7^3 \)
Answer: To find \( 7^3 \) as a sum of consecutive odd numbers, we follow the pattern from Question 6. The first odd number for \( n^3 \) is \( n(n-1) + 1 \). For \( n=7 \), the first odd number is \( 7(7-1) + 1 = 7 \times 6 + 1 = 42 + 1 = 43 \). We then need to add 7 consecutive odd numbers starting from 43.
\( 7^3 = 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343 \). Each cube has a unique set of consecutive odd numbers that sum up to it.
In simple words: For \( 7^3 \), we start from the odd number 43 and add the next six odd numbers after it. This sum of seven odd numbers equals 343.

🎯 Exam Tip: Always make sure you are adding the correct number of consecutive odd numbers (equal to 'n' for \( n^3 \)) and that you start with the correct first odd number for that 'n'.

Question 8. Fill in the blanks
Answer:

This pattern shows that when you find the cube of a number that is a product of other factors, you can cube each factor separately and then multiply them. This is a property of exponents, \( (ab)^n = a^n b^n \). Each prime factor in the cube of a number will appear three times, or in multiples of three.
In simple words: When you cube a number that is made by multiplying other numbers (like \( 6 = 2 \times 3 \)), you can cube each of those smaller numbers and then multiply their cubes together to get the final answer.

🎯 Exam Tip: Remember the exponent rule \( (a \times b)^3 = a^3 \times b^3 \). This allows you to break down a larger cube into the cubes of its prime factors, simplifying calculations.

Question 9. Check the perfect cube in the following numbers?
(i) 2700
(ii) 16000
(iii) 64000
(iv) 900
(v) 125000
(vi) 36000
Answer: A number is a perfect cube if all its prime factors can be grouped into sets of three. We will find the prime factorization for each number.
(i) 2700
Prime factors of \( 2700 = 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 = 2^2 \times 3^3 \times 5^2 \).
Here, the prime factors 2 and 5 do not appear in groups of three.
Thus, 2700 is not a perfect cube.

(ii) 16000
Prime factors of \( 16000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^7 \times 5^3 = (2^3 \times 2^3 \times 2) \times 5^3 \).
Here, the prime factor 2 does not appear completely in groups of three (one 2 is left).
Thus, 16000 is not a perfect cube.

(iii) 64000
Prime factors of \( 64000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 = 2^9 \times 5^3 = (2^3 \times 2^3 \times 2^3) \times 5^3 \).
Here, all prime factors (2 and 5) appear in perfect groups of three.
Thus, 64000 is a perfect cube. (It is \( 40^3 \)).

(iv) 900
Prime factors of \( 900 = 2 \times 2 \times 3 \times 3 \times 5 \times 5 = 2^2 \times 3^2 \times 5^2 \).
Here, none of the prime factors (2, 3, and 5) appear in groups of three.
Thus, 900 is not a perfect cube.

(v) 125000
Prime factors of \( 125000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 2^3 \times 5^6 = 2^3 \times (5^3 \times 5^3) \).
Here, all prime factors (2 and 5) appear in perfect groups of three.
Thus, 125000 is a perfect cube. (It is \( 50^3 \)).

(vi) 36000
Prime factors of \( 36000 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5 = 2^5 \times 3^2 \times 5^3 = (2^3 \times 2^2) \times 3^2 \times 5^3 \).
Here, the prime factors 2 and 3 do not appear completely in groups of three.
Thus, 36000 is not a perfect cube.

(vii) 21600
Prime factors of \( 21600 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 = 2^5 \times 3^3 \times 5^2 = (2^3 \times 2^2) \times 3^3 \times 5^2 \).
Here, the prime factors 2 and 5 do not appear completely in groups of three.
Thus, 21600 is not a perfect cube.

(viii) 10000
Prime factors of \( 10000 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = 2^4 \times 5^4 = (2^3 \times 2) \times (5^3 \times 5) \).
Here, the prime factors 2 and 5 do not appear completely in groups of three.
Thus, 10000 is not a perfect cube.

(ix) 27000000
Prime factors of \( 27000000 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 2^6 \times 3^3 \times 5^6 = (2^3 \times 2^3) \times 3^3 \times (5^3 \times 5^3) \).
Here, all prime factors (2, 3, and 5) appear in perfect groups of three.
Thus, 27000000 is a perfect cube. (It is \( 300^3 \)).

(x) 11000
Prime factors of \( 11000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 11 = 2^3 \times 5^3 \times 11^1 \).
Here, the prime factor 11 does not appear in a group of three.
Thus, 11000 is not a perfect cube.
In simple words: A number is a perfect cube if, when you break it down into its smallest prime factors, every single prime factor appears exactly three times, or in groups of three (like \( 2 \times 2 \times 2 \)). If any factor is left over, it's not a perfect cube.

🎯 Exam Tip: To determine if a number is a perfect cube, always use prime factorization. Group the prime factors in sets of three. If all factors form complete triplets, it's a perfect cube; otherwise, it's not.