GSEB Class 8 Maths Solutions Chapter 7 Cube and Cube Roots InText Questions

Try These (Page 111)

Question 1. Find the ones digit of the cube of each of the following numbers?
1. 3331
2. 8888
3. 149
4. 1005
5. 1024
6. 77
7. 5022
8. 53
Answer:

In simple words: To find the units digit of a cube, just look at the units digit of the original number and find its cube. The units digit of that cube is your answer.

Exam Tip: Remember the pattern of unit digits for cubes: 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9, 0→0. Only 2, 3, 7, 8 have different unit digits from their original numbers.

Note:

1. There are only 4 numbers, less than 100 which are perfect cubes. They are:
1, 8, 27 and 64.

2. There are only 5 numbers between 100 and 1000 which are perfect cubes.
These are: 125, 216, 343, 512 and 729.

3. The least 4 digit perfect cube is 1000.

Try These (Page 111)

Question 1. Consider the following pattern:
\( 1^{3} = 1 \)
\( 3 + 5 = 8 = 2^{3} \)
\( 7 + 9 + 11 = 27 = 3^{3} \)
\( 13 + 15 + 17 + 19 = 64 = 4^{3} \)
\( 21 + 23 + 25 + 27 + 29 = 125 = 5^{2} \)
Express the following numbers as the sum of odd numbers using the above pattern?
(a) \( 6^{3} \)
(b) \( 8^{3} \)
(c) \( 7^{3} \)
Answer:
(a) For \( 6^{3} \): Here, \( n = 6 \) and \( (n - 1) = 5 \).
We start with \( (6 \times 5) + 1 = 30 + 1 = 31 \).
So, we have: \( 6^{3} = 31 + 33 + 35 + 37 + 39 + 41 \)
\( = 216 \)
(b) For \( 8^{3} \): Here, \( n = 8 \) and \( (n - 1) = 7 \).
We start with \( (8 \times 7) + 1 = 56 + 1 = 57 \).
So, we have: \( 8^{3} = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 \)
\( = 512 \)
(c) For \( 7^{3} \): Here, \( n = 7 \) and \( (n - 1) = 6 \).
We start with \( (7 \times 6) + 1 = 42 + 1 = 43 \).
So, we have: \( 7^{3} = 43 + 45 + 47 + 49 + 51 + 53 + 55 \)
\( = 343 \)
In simple words: This pattern shows that you can write any cube as a sum of consecutive odd numbers. For \( n^3 \), start with the odd number \( (n \times (n-1)) + 1 \) and add \( n \) consecutive odd numbers from there.

Exam Tip: To find the starting odd number for \( n^3 \), calculate \( (n \times (n-1)) + 1 \). Then, list \( n \) consecutive odd numbers beginning with this value to get the sum.

Question 2. Consider the following pattern:
\( 2^{3}-1^{3} = 1 + 2 \times 1 \times 3 \)
\( 3^{3}-2^{3} = 1 + 3 \times 2 \times 3 \)
\( 4^{3}-3^{3} = 1 + 4 \times 3 \times 3 \)
Using the above pattern, find the value of the following:
1. \( 7^{3}-6^{3} \)
2. \( 12^{3}-11^{3} \)
3. \( 20^{3}-19^{3} \)
4. \( 51^{3}-50^{3} \)
Answer:
This pattern indicates that \( n^{3} - (n-1)^{3} = 1 + n \times (n-1) \times 3 \).
1. \( 7^{3}-6^{3} \)
Using the pattern, we get: \( 1 + 7 \times 6 \times 3 \)
\( = 1 + 126 \)
\( = 127 \)
2. \( 12^{3}-11^{3} \)
Using the pattern, we get: \( 1 + 12 \times 11 \times 3 \)
\( = 1 + 396 \)
\( = 397 \)
3. \( 20^{3}-19^{3} \)
Using the pattern, we get: \( 1 + 20 \times 19 \times 3 \)
\( = 1 + 1140 \)
\( = 1141 \)
4. \( 51^{3}-50^{3} \)
Using the pattern, we get: \( 1 + 51 \times 50 \times 3 \)
\( = 1 + 7650 \)
\( = 7651 \)
In simple words: The difference between two consecutive cubes, \( n^{3} \) and \( (n-1)^{3} \), can be easily calculated. You just need to take 1, then add the product of \( n \), \( (n-1) \), and 3.

Exam Tip: This pattern is a special case of the algebraic identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). For consecutive numbers where \( a = n \) and \( b = n-1 \), \( a-b = 1 \), simplifying the formula to \( n^2 + n(n-1) + (n-1)^2 \). This simplifies further to \( 1 + 3n(n-1) \) which is the given pattern.

Note:

In the prime factorisation of any number, if each factor appears three times, then the number is a perfect cube.

Try These (Page 112)

Question 1. Which of the following are perfect cubes?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Answer:
1. For 400:
Prime factors of 400 are \( 2 \times 2 \times 2 \times 2 \times 5 \times 5 \).
When we group these factors into triples, we find that \( 2 \times 5 \times 5 \) are left over without forming a complete group of three.
Therefore, 400 is not a perfect cube.
2. For 3375:
Prime factors of 3375 are \( 3 \times 3 \times 3 \times 5 \times 5 \times 5 \).
We can group these prime factors into triples: \( (3 \times 3 \times 3) \times (5 \times 5 \times 5) \). No factors are left over.
Therefore, 3375 is a perfect cube.
3. For 8000:
Prime factors of 8000 are \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \).
We can group these prime factors into triples: \( (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) \). No factors are left over.
Therefore, 8000 is a perfect cube.
4. For 15625:
Prime factors of 15625 are \( 5 \times 5 \times 5 \times 5 \times 5 \times 5 \).
We can group these prime factors into triples: \( (5 \times 5 \times 5) \times (5 \times 5 \times 5) \). No factors are left over.
Therefore, 15625 is a perfect cube.
5. For 9000:
Prime factors of 9000 are \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5 \).
When we group these factors into triples, we find that \( 3 \times 3 \) are left over without forming a complete group of three.
Therefore, 9000 is not a perfect cube.
6. For 6859:
Prime factors of 6859 are \( 19 \times 19 \times 19 \).
We can group these prime factors into triples: \( (19 \times 19 \times 19) \). No factors are left over.
Therefore, 6859 is a perfect cube.
7. For 2025:
Prime factors of 2025 are \( 3 \times 3 \times 3 \times 3 \times 5 \times 5 \).
When we group these factors into triples, we find that \( 3 \times 5 \times 5 \) are left over without forming a complete group of three.
Therefore, 2025 is not a perfect cube.
8. For 10648:
Prime factors of 10648 are \( 2 \times 2 \times 2 \times 11 \times 11 \times 11 \).
We can group these prime factors into triples: \( (2 \times 2 \times 2) \times (11 \times 11 \times 11) \). No factors are left over.
Therefore, 10648 is a perfect cube.
In simple words: To check if a number is a perfect cube, you need to break it down into its prime factors. If you can make groups of three identical factors for every prime number, then it is a perfect cube. If any factors are left alone, it's not a perfect cube.

Exam Tip: Always perform prime factorization systematically. For a number to be a perfect cube, every prime factor in its factorization must appear a multiple of three times (i.e., in groups of 3).

Try These (Page 113)

Question 1. Check which of the following are perfect cubes?
(i) 2700
(ii) 16000
(iii) 64000
(iv) 900
(v) 125000
(vi) 36000
(vii) 21600
(viii) 10000
(ix) 27000000
(x) 1000
What pattern do you observe in these perfect cubes?
Answer:
(i) For 2700:
Prime factors of 2700 are \( 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \).
We do not get complete triples of prime factors, as \( 2 \times 2 \) and \( 5 \times 5 \) are left over.
Therefore, 2700 is not a perfect cube.
(ii) For 16000:
Prime factors of 16000 are \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \).
After grouping in threes, we get \( (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) \). No factors are left over.
Therefore, 16000 is a perfect cube.
(iii) For 64000:
Prime factors of 64000 are \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \).
Since we get groups of triples for all prime factors: \( (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (5 \times 5 \times 5) \).
Therefore, 64000 is a perfect cube.
(iv) For 900:
Prime factors of 900 are \( 2 \times 2 \times 3 \times 3 \times 5 \times 5 \).
These factors are ungrouped in triples; none of the factors appear three times.
Therefore, 900 is not a perfect cube.
(v) For 125000:
Prime factors of 125000 are \( 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \).
As we get all the prime factors in groups of triples: \( (2 \times 2 \times 2) \times (5 \times 5 \times 5) \times (5 \times 5 \times 5) \).
Therefore, 125000 is a perfect cube.
(vi) For 36000:
Prime factors of 36000 are \( 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5 \).
While grouping the prime factors of 36000 in triples, we are left over with \( 2 \times 2 \) and \( 3 \times 3 \).
Therefore, 36000 is not a perfect cube.
(vii) For 21600:
Prime factors of 21600 are \( 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \).
While grouping the prime factors of 21600 in triples, we are left with \( 2 \times 2 \) and \( 5 \times 5 \).
Therefore, 21600 is not a perfect cube.
(viii) For 10000:
Prime factors of 10000 are \( 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 \).
While grouping the prime factors into triples, we are left over with \( 2 \) and \( 5 \).
Therefore, 10000 is not a perfect cube.
(ix) For 27000000:
Prime factors of 27000000 are \( 2^6 \times 3^3 \times 5^6 \).
Since all the prime factors of 27000000 appear in groups of triples, \( (2^3 \times 2^3 \times 3^3 \times 5^3 \times 5^3) \).
Therefore, 27000000 is a perfect cube.
(x) For 1000:
Prime factors of 1000 are \( 2 \times 2 \times 2 \times 5 \times 5 \times 5 \).
Since all the prime factors of 1000 appear in groups of triples: \( (2 \times 2 \times 2) \times (5 \times 5 \times 5) \).
Therefore, 1000 is a perfect cube.

**Pattern Observation:**
Now, in these perfect cubes, we can say that “the number of zeros at the end of a perfect cube must be 3 or a multiple of 3, failing which the number cannot be a perfect cube.”
In simple words: We checked each number by breaking it down into prime factors and seeing if all factors could be grouped in sets of three. Perfect cubes like 16000, 64000, 125000, 27000000, and 1000 had all factors in triples. Numbers like 2700, 900, 36000, 21600, and 10000 had some factors left out, so they are not perfect cubes. The pattern we noticed is that perfect cubes always have a number of zeros at the end that is a multiple of three.

Exam Tip: A quick check for perfect cubes ending in zeros is to count the zeros. If the count of zeros is not a multiple of three (e.g., 1, 2, 4, 5 zeros), the number cannot be a perfect cube. Then proceed with prime factorization for a definitive answer.

Try These (Page 115)

Question 1. State true or false: for any integer m, \( m^{2} < m^{3} \). Why?
Answer:
This statement is not always true.
For instance, let \( m = -1 \).
Then, \( m^{2} = (-1)^{2} = 1 \).
And \( m^{3} = (-1)^{3} = -1 \).
In this case, \( 1 \) is not less than \( -1 \), so \( m^{2} < m^{3} \) is false for \( m = -1 \).
It is also false for \( m = 0 \) (\( 0^2 < 0^3 \) means \( 0 < 0 \), which is false) and for \( m = 1 \) (\( 1^2 < 1^3 \) means \( 1 < 1 \), which is false).
Therefore, the statement \( m^{2} < m^{3} \) is false.
In simple words: The statement that \( m^{2} \) is always smaller than \( m^{3} \) for any integer \( m \) is incorrect. For example, if \( m \) is a negative number like -1, then \( m^{2} \) becomes 1 and \( m^{3} \) becomes -1, and 1 is not smaller than -1. It is also not true for 0 or 1.

Exam Tip: When evaluating "true or false" statements involving variables, always test with various types of numbers: positive integers, negative integers, zero, and fractions to find a counterexample if one exists.