Official NCERT 2026-27 Solutions
NCERT Solutions for Class 8 Maths Chapter 1: A Square and A Cube (Ganita Prakash)
Comprehensive, step-by-step handwritten solutions in authentic Topper Notebook style. Covers every textbook exploration, in-text investigation, and “Figure it Out” exercise from Pages 2 to 17 of Ganita Prakash Part I.
Quick Chapter Navigation & Blueprint
- Part A: Properties of Square Numbers (Pages 2–7) — Factor parity, locker puzzle, unit digit patterns, zeros rule, and consecutive square gaps.
- Part B: “Figure it Out” Section 1.1 (Page 10) — Questions 1 to 8: Testing squareness, last digits, algebraic identities, area of squares, LCM & square numbers, prime factorisation, and algebraic number patterns.
- Part C: Cubes and Cube Roots Explorations (Pages 13–15) — Trailing zeroes in cubes, consecutive odd number sums, and direct cube root identification.
- Part D: “Figure it Out” Section 1.2 (Pages 16–17) — Questions 1 to 5: Prime factorisation for cube roots, finding missing multipliers, true/false analysis with proofs, estimation method, and difference of powers comparison.
Official Textbook PDF Reference: Download NCERT Class 8 Maths Chapter 1 (hegp101.pdf)
Part A: In-Text Explorations & Foundations (Pages 2–7)
Exploration 1 (Page 2):
(a) Does every number have an even number of factors?
(b) Can you use this insight to find more numbers with an odd number of factors?
(a) Does every number have an even number of factors?
(b) Can you use this insight to find more numbers with an odd number of factors?
Topper’s Step-by-Step Derivation & Proof:
- Factor Pairing Analysis: Every factor of an integer N comes in pairs: if a is a factor of N, then there is a paired factor b such that a × b = N.
- For example, for the number 12: factors are (1, 12), (2, 6), (3, 4). Since a ≠ b in all pairs, the total number of factors is 3 pairs × 2 = 6 (which is even).
- The Exceptional Case (Perfect Squares): If a number is a perfect square, N = k2 = k × k. In this case, one of the pairs consists of identical factors (k, k). When listing distinct factors, k is counted only once!
- For example, for 36: pairs are (1, 36), (2, 18), (3, 12), (4, 9), and the solitary pair (6, 6). The distinct factors are {1, 2, 3, 4, 6, 9, 12, 18, 36}, giving exactly 9 factors (an odd number).
Conclusion:
(a) No, not every number has an even number of factors.
(b) Yes, all perfect square numbers (and ONLY perfect square numbers) have an odd number of distinct factors! Examples: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, etc.
(a) No, not every number has an even number of factors.
(b) Yes, all perfect square numbers (and ONLY perfect square numbers) have an odd number of distinct factors! Examples: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, etc.
Exam Tip: A positive integer N has an odd number of factors if and only if N is a perfect square. This fundamental theorem frequently appears in competitive Olympiads and school exams!
Exploration 2 (Page 3 – The 100 Locker Problem):
There are 100 lockers numbered 1 to 100 in a school hallway. Initially, all lockers are closed. 100 students walk by in sequence:
– Student 1 toggles every locker (opens all).
– Student 2 toggles every 2nd locker (closes 2, 4, 6, 8…).
– Student 3 toggles every 3rd locker (toggles 3, 6, 9, 12…).
– And so on, until Student 100 toggles locker 100.
Which locker numbers will remain open at the very end?
There are 100 lockers numbered 1 to 100 in a school hallway. Initially, all lockers are closed. 100 students walk by in sequence:
– Student 1 toggles every locker (opens all).
– Student 2 toggles every 2nd locker (closes 2, 4, 6, 8…).
– Student 3 toggles every 3rd locker (toggles 3, 6, 9, 12…).
– And so on, until Student 100 toggles locker 100.
Which locker numbers will remain open at the very end?
Topper’s Mathematical Analysis:
- Locker Toggle Logic: A locker numbered k is toggled once by every student whose number is a divisor (factor) of k.
- If locker k has an even number of divisors, it gets toggled an even number of times: Open → Closed → Open → Closed… ending up CLOSED.
- If locker k has an odd number of divisors, it gets toggled an odd number of times: Open → Closed → Open… ending up OPEN.
- From our proof in Exploration 1, only perfect square numbers have an odd number of divisors.
Lockers that remain open: All square numbers between 1 and 100.
1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. (Total 10 lockers remain open).
1, 4, 9, 16, 25, 36, 49, 64, 81, and 100. (Total 10 lockers remain open).
Exploration 3 (Page 5 – Units Digit of Squares):
Which of the following numbers have the digit 6 in the units place?
(i) 382 (ii) 342 (iii) 462 (iv) 562 (v) 742 (vi) 822
Which of the following numbers have the digit 6 in the units place?
(i) 382 (ii) 342 (iii) 462 (iv) 562 (v) 742 (vi) 822
Rule of Units Digit in Squares:
- The units digit of a square number depends solely on the units digit of the base number:
• Base ends in 4 → 4 × 4 = 16 → Square ends in 6.
• Base ends in 6 → 6 × 6 = 36 → Square ends in 6.
• Base ends in 8 → 8 × 8 = 64 → Square ends in 4.
• Base ends in 2 → 2 × 2 = 4 → Square ends in 4. - Testing each given number:
(i) 38 ends in 8 → 382 ends in 4.
(ii) 34 ends in 4 → 342 ends in 6.
(iii) 46 ends in 6 → 462 ends in 6.
(iv) 56 ends in 6 → 562 ends in 6.
(v) 74 ends in 4 → 742 ends in 6.
(vi) 82 ends in 2 → 822 ends in 4.
Final Answer: Numbers having units digit 6 are (ii) 342, (iii) 462, (iv) 562, and (v) 742.
Exploration 4 (Page 5 – The Zeros Rule for Squares):
(a) If a number contains 3 zeros at the end, how many zeros will its square have at the end?
(b) What do you notice about the number of zeros at the end of a number and its square? Will this always happen?
(c) Can we say that square numbers can only have an even number of zeros at the end?
(a) If a number contains 3 zeros at the end, how many zeros will its square have at the end?
(b) What do you notice about the number of zeros at the end of a number and its square? Will this always happen?
(c) Can we say that square numbers can only have an even number of zeros at the end?
Detailed Proof & Analysis:
- Let the number be expressed in scientific/algebraic notation as N = k × 10m, where k is not divisible by 10, and m is the number of trailing zeros.
- Squaring both sides gives: N2 = (k × 10m)2 = k2 × 102m.
- Because k does not end in zero, k2 cannot produce any extra zeros. Therefore, the square contains exactly 2m zeros (double the original number of zeros).
- Since 2m is always a multiple of 2, a square number can never end in an odd number of zeros (such as 10, 1000, 100000).
Answers:
(a) A number with 3 zeros at the end will have 3 × 2 = 6 zeros at the end of its square.
(b) The number of trailing zeros in a square is always double the number of trailing zeros in the original number. Yes, this always happens.
(c) Yes, a perfect square can only end with an even number of zeros (e.g., 2, 4, 6 zeros).
(a) A number with 3 zeros at the end will have 3 × 2 = 6 zeros at the end of its square.
(b) The number of trailing zeros in a square is always double the number of trailing zeros in the original number. Yes, this always happens.
(c) Yes, a perfect square can only end with an even number of zeros (e.g., 2, 4, 6 zeros).
Exploration 5 (Page 7 – Numbers Between Consecutive Squares):
Find how many non-square numbers lie between two consecutive perfect squares n2 and (n + 1)2. Do you notice a pattern?
Find how many non-square numbers lie between two consecutive perfect squares n2 and (n + 1)2. Do you notice a pattern?
Algebraic Formula & Step-by-Step Derivation:
- Let the two consecutive integers be n and n + 1.
- Their squares are n2 and (n + 1)2 = n2 + 2n + 1.
- The count of numbers strictly lying between them is:
Count = [(n + 1)2 − n2] − 1
= [n2 + 2n + 1 − n2] − 1
= 2n + 1 − 1 = 2n. - Verification with small values:
• Between 12 (1) and 22 (4): numbers are 2, 3 → exactly 2 × 1 = 2 numbers.
• Between 22 (4) and 32 (9): numbers are 5, 6, 7, 8 → exactly 2 × 2 = 4 numbers.
• Between 32 (9) and 42 (16): numbers are 10, 11, 12, 13, 14, 15 → exactly 2 × 3 = 6 numbers.
Formula: Between n2 and (n + 1)2, there are always 2n non-square numbers.
Part B: “Figure it Out” Section 1.1 (Page 10)
Question 1 (Page 10):
Which of the following numbers are not perfect squares?
(i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Which of the following numbers are not perfect squares?
(i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Topper’s Elimination Method via Unit Digits:
- Golden Rule of Perfect Squares: The square of any whole number can only end in the digits 0, 1, 4, 5, 6, or 9. A number ending in 2, 3, 7, or 8 is NEVER a perfect square!
- (i) 2032: Ends in digit 2. Hence, it CANNOT be a perfect square.
- (ii) 2048: Ends in digit 8. Hence, it CANNOT be a perfect square.
- (iii) 1027: Ends in digit 7. Hence, it CANNOT be a perfect square.
- (iv) 1089: Ends in digit 9 (valid square ending). By prime factorisation: 1089 = 3 × 3 × 11 × 11 = (3 × 11)2 = 332. Thus, 1089 is a perfect square!
Final Answer: The numbers that are not perfect squares are (i) 2032, (ii) 2048, and (iii) 1027.
Question 2 (Page 10):
Which one among 642, 1082, 2922, 362 has last digit 4?
Which one among 642, 1082, 2922, 362 has last digit 4?
Step-by-Step Unit Digit Evaluation:
- 64 ends in 4 → 4 × 4 = 16 → 642 ends in 6.
- 108 ends in 8 → 8 × 8 = 64 → 1082 ends in 4.
- 292 ends in 2 → 2 × 2 = 4 → 2922 ends in 4.
- 36 ends in 6 → 6 × 6 = 36 → 362 ends in 6.
Final Answer: Both 1082 and 2922 have last digit 4.
Question 3 (Page 10):
Given 1252 = 15625, what is the value of 1262?
(i) 15625 + 126 (ii) 15625 + 262 (iii) 15625 + 253
(iv) 15625 + 251 (v) 15625 + 512
Given 1252 = 15625, what is the value of 1262?
(i) 15625 + 126 (ii) 15625 + 262 (iii) 15625 + 253
(iv) 15625 + 251 (v) 15625 + 512
Algebraic Identity Derivation:
- We express 126 as (125 + 1).
- Applying the expansion identity (a + 1)2 = a2 + 2a + 1 = a2 + [a + (a + 1)]:
- Here a = 125. Therefore:
1262 = (125 + 1)2
= 1252 + 2(125) + 1
= 15625 + 250 + 1
= 15625 + 251 = 15876.
Correct Option: (iv) 15625 + 251.
Question 4 (Page 10):
Find the length of the side of a square whose area is 441 m2.
Find the length of the side of a square whose area is 441 m2.
Step-by-Step Geometric Calculation:
- Let the side length of the square be s metres.
- Area of a square = s2 = 441 m2.
- Side length s = √441.
- Prime factorisation of 441:
441 ÷ 3 = 147
147 ÷ 3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1
441 = (3 × 3) × (7 × 7) = 32 × 72 - Taking square root: s = 3 × 7 = 21 m.
Final Answer: The length of the side of the square is 21 metres.
Question 5 (Page 10):
Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Two-Step LCM & Pairing Method:
- Step 1: Find the LCM of 4, 9, and 10:
• Prime factors of 4 = 2 × 2 = 22
• Prime factors of 9 = 3 × 3 = 32
• Prime factors of 10 = 2 × 5
LCM(4, 9, 10) = 22 × 32 × 5 = 4 × 9 × 5 = 180. - Step 2: Make the LCM a perfect square by pairing up prime factors:
180 = (2 × 2) × (3 × 3) × 5.
Notice that prime factor 5 is unpaired! - To get the smallest multiple that is a perfect square, we must multiply 180 by 5:
Required square number = 180 × 5 = 900.
(Check: 900 = 302, divisible by 4 (225), 9 (100), and 10 (90)).
Final Answer: The smallest square number divisible by 4, 9, and 10 is 900.
Question 6 (Page 10):
Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Step-by-Step Prime Factorisation:
- Dividing 9408 successively by prime numbers:
9408 ÷ 2 = 4704
4704 ÷ 2 = 2352
2352 ÷ 2 = 1176
1176 ÷ 2 = 588
588 ÷ 2 = 294
294 ÷ 2 = 147
147 ÷ 3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1 - Grouping into prime pairs:
9408 = (2 × 2) × (2 × 2) × (2 × 2) × 3 × (7 × 7)
= 26 × 31 × 72. - The prime factor 3 is unpaired. Therefore, the smallest number by which 9408 must be multiplied is 3.
- New Product = 9408 × 3 = 28224.
- Square Root of product = √28224 = 23 × 3 × 7 = 8 × 21 = 168.
Final Answer:
Smallest multiplying number = 3
Square root of the product = 168
Smallest multiplying number = 3
Square root of the product = 168
Question 7 (Page 10):
How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100
How many numbers lie between the squares of the following numbers?
(i) 16 and 17
(ii) 99 and 100
Application of the 2n Formula:
- From our derivation, the number of non-square integers lying between n2 and (n + 1)2 is given by 2n.
- (i) Between 162 and 172:
Here n = 16.
Number of integers = 2 × 16 = 32.
(Check: 172 − 162 − 1 = 289 − 256 − 1 = 32). - (ii) Between 992 and 1002:
Here n = 99.
Number of integers = 2 × 99 = 198.
(Check: 1002 − 992 − 1 = 10000 − 9801 − 1 = 198).
Final Answers: (i) 32 numbers | (ii) 198 numbers.
Question 8 (Page 10 – Number Pattern):
In the following pattern, fill in the missing numbers:
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + 202 = (…)2
92 + 102 + (…)2 = (91)2
In the following pattern, fill in the missing numbers:
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + 202 = (…)2
92 + 102 + (…)2 = (91)2
Pattern Discovery & Algebraic Proof:
- Observing the columns for each row n:
• 1st term = n2
• 2nd term = (n + 1)2
• 3rd term = [n(n + 1)]2 (the product of the base of the first two numbers!)
• Right hand side = [n(n + 1) + 1]2 (one more than the 3rd base!) - Algebraic Proof:
n2 + (n + 1)2 + [n(n + 1)]2
= n2 + n2 + 2n + 1 + n2(n + 1)2
= n2(n + 1)2 + 2n(n + 1) + 1
= [n(n + 1) + 1]2 — which confirms the pattern perfectly! - Row 4: 42 + 52 + 202 → RHS base = 20 + 1 = 21 → (21)2.
- Row 5: 92 + 102 + (…)2 = (91)2 → 3rd base = 9 × 10 = 90 → (90)2.
Completed Pattern:
42 + 52 + 202 = (21)2
92 + 102 + (90)2 = (91)2
42 + 52 + 202 = (21)2
92 + 102 + (90)2 = (91)2
Part C: Cube Explorations (Pages 13–15)
Exploration 6 (Page 13):
Can a cube end with exactly two zeroes (00)? Explain with mathematical proof.
Can a cube end with exactly two zeroes (00)? Explain with mathematical proof.
Topper’s Mathematical Proof:
- Let a natural number N end with m zeros, represented as N = k × 10m, where k is not a multiple of 10.
- Cubing both sides gives: N3 = (k × 10m)3 = k3 × 103m.
- Since k does not end in zero, k3 does not contribute any additional trailing zeros.
- Therefore, the total number of zeros at the end of a perfect cube is always 3m (a multiple of 3).
- Hence, a perfect cube can end with 0, 3, 6, 9… zeros, but never with 1, 2, 4, 5 zeros.
Conclusion: No, a perfect cube can never end with exactly two zeros.
Exploration 7 (Page 14):
Can you tell what this sum of consecutive odd numbers is without doing manual addition?
91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109 = ?
Can you tell what this sum of consecutive odd numbers is without doing manual addition?
91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109 = ?
Consecutive Odd Numbers Cube Property:
- Recall the beautiful relationship between consecutive odd numbers and cubes:
Row 1: 1 = 13 = 1 (1 number)
Row 2: 3 + 5 = 23 = 8 (2 numbers)
Row 3: 7 + 9 + 11 = 33 = 27 (3 numbers)
Row 4: 13 + 15 + 17 + 19 = 43 = 64 (4 numbers) - Here, there are exactly 10 odd numbers: 91, 93, 95, 97, 99, 101, 103, 105, 107, 109.
- The first number in row 10 is 10 × 9 + 1 = 91. The last is 109.
- Therefore, by the sum-of-odd-numbers property, the sum of these 10 numbers equals 103!
Sum = 103 = 1000.
Exploration 8 (Page 15):
Find the cube roots of these numbers:
(i) ∛64 (ii) ∛512 (iii) ∛729
Find the cube roots of these numbers:
(i) ∛64 (ii) ∛512 (iii) ∛729
Direct Cube Root Identification:
- (i) ∛64: Since 4 × 4 × 4 = 43 = 64 → ∛64 = 4.
- (ii) ∛512: Since 8 × 8 × 8 = 83 = 512 → ∛512 = 8.
- (iii) ∛729: Since 9 × 9 × 9 = 93 = 729 → ∛729 = 9.
Final Answers: (i) 4 | (ii) 8 | (iii) 9.
Part D: “Figure it Out” Section 1.2 (Pages 16–17)
Question 1 (Page 16):
Find the cube roots of 27000 and 10648 by prime factorisation.
Find the cube roots of 27000 and 10648 by prime factorisation.
Topper’s Prime Factorisation & Triplet Grouping:
- (i) For 27000:
27000 = 27 × 1000 = (3 × 3 × 3) × (10 × 10 × 10)
In complete prime factorisation:
27000 = (2 × 2 × 2) × (3 × 3 × 3) × (5 × 5 × 5)
= 23 × 33 × 53
Taking cube root: ∛27000 = 2 × 3 × 5 = 30. - (ii) For 10648:
10648 ÷ 2 = 5324
5324 ÷ 2 = 2662
2662 ÷ 2 = 1331
1331 ÷ 11 = 121
121 ÷ 11 = 11
11 ÷ 11 = 1
10648 = (2 × 2 × 2) × (11 × 11 × 11) = 23 × 113.
Taking cube root: ∛10648 = 2 × 11 = 22.
Final Answer:
(i) Cube root of 27000 = 30
(ii) Cube root of 10648 = 22
(i) Cube root of 27000 = 30
(ii) Cube root of 10648 = 22
Question 2 (Page 16):
What smallest number will you multiply by 1323 to make it a perfect cube number?
What smallest number will you multiply by 1323 to make it a perfect cube number?
Prime Factorisation & Triplet Analysis:
- Dividing 1323 by prime factors:
1323 ÷ 3 = 441
441 ÷ 3 = 147
147 ÷ 3 = 49
49 ÷ 7 = 7
7 ÷ 7 = 1 - Grouping into prime factors:
1323 = (3 × 3 × 3) × (7 × 7) = 33 × 72. - Notice that prime factor 3 forms a complete triplet (33), but prime factor 7 only has 2 factors (72).
- To complete the triplet of 7s, we need one more 7.
- Multiplying by 7: 1323 × 7 = 9261 = 33 × 73 = (3 × 7)3 = 213.
Final Answer: The smallest number to multiply by 1323 to make it a perfect cube is 7.
Question 3 (Page 16):
State True or False. Explain your reasoning with proof or counterexample.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
State True or False. Explain your reasoning with proof or counterexample.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
Topper’s Rigorous Reasoning:
- (i) The cube of any odd number is even:
FALSE. The product of odd numbers is always odd: (odd) × (odd) × (odd) = odd. For example, 33 = 27 (odd), 53 = 125 (odd). - (ii) There is no perfect cube that ends with 8:
FALSE. The cube of any number ending in 2 ends in 8! For example, 23 = 8, 123 = 1728, 223 = 10648. All of these end in 8. - (iii) The cube of a 2-digit number may be a 3-digit number:
FALSE. The smallest 2-digit number is 10. Its cube is 103 = 1000, which has 4 digits. Hence, no 2-digit number can have a cube with fewer than 4 digits. - (iv) The cube of a 2-digit number may have seven or more digits:
FALSE. The largest 2-digit number is 99. Its cube is 993 = 970299, which has exactly 6 digits. Therefore, the cube of any 2-digit number can only have 4, 5, or 6 digits — never 7 or more. - (v) Cube numbers have an odd number of factors:
FALSE. Only perfect squares have an odd number of factors. A perfect cube can have an even number of factors. For example, 8 = 23 has factors {1, 2, 4, 8} (total 4 factors, which is even). 27 = 33 has factors {1, 3, 9, 27} (total 4 factors, even).
Summary:
(i) False | (ii) False | (iii) False | (iv) False | (v) False.
(i) False | (ii) False | (iii) False | (iv) False | (v) False.
Question 4 (Page 16 – The Estimation Method):
You are told that 1331 is a perfect cube. Guess without factorisation what its cube root is. Similarly, guess the cube roots of 4913, 12167, and 32768.
You are told that 1331 is a perfect cube. Guess without factorisation what its cube root is. Similarly, guess the cube roots of 4913, 12167, and 32768.
Topper’s Rapid 2-Group Estimation Technique:
- Units Digit Cube Mapping:
Ends in 1 → root ends in 1 | Ends in 3 → root ends in 7 | Ends in 7 → root ends in 3 | Ends in 8 → root ends in 2. - For 1331:
• Group 1 (last 3 digits): 331 → ends in 1, so units digit of root is 1.
• Group 2 (remaining digits): 1 → 13 ≤ 1 < 23, so tens digit is 1.
→ ∛1331 = 11. - For 4913:
• Group 1: 913 → ends in 3, so units digit is 7 (since 73 = 343).
• Group 2: 4 → 13 (1) ≤ 4 < 23 (8), so tens digit is 1.
→ ∛4913 = 17. - For 12167:
• Group 1: 167 → ends in 7, so units digit is 3 (since 33 = 27).
• Group 2: 12 → 23 (8) ≤ 12 < 33 (27), so tens digit is 2.
→ ∛12167 = 23. - For 32768:
• Group 1: 768 → ends in 8, so units digit is 2 (since 23 = 8).
• Group 2: 32 → 33 (27) ≤ 32 < 43 (64), so tens digit is 3.
→ ∛32768 = 32.
Estimated Cube Roots:
∛1331 = 11 | ∛4913 = 17 | ∛12167 = 23 | ∛32768 = 32.
∛1331 = 11 | ∛4913 = 17 | ∛12167 = 23 | ∛32768 = 32.
Question 5 (Page 17):
Which of the following is the greatest? Explain your reasoning.
(i) 673 − 663
(ii) 433 − 423
(iii) 672 − 662
(iv) 432 − 422
Which of the following is the greatest? Explain your reasoning.
(i) 673 − 663
(ii) 433 − 423
(iii) 672 − 662
(iv) 432 − 422
Topper’s Algebraic Evaluation & Comparison:
- Comparing Difference of Squares:
(iii) 672 − 662 = (67 − 66)(67 + 66) = 1 × 133 = 133.
(iv) 432 − 422 = (43 − 42)(43 + 42) = 1 × 85 = 85. - Comparing Difference of Cubes using Identity a3 − b3 = (a − b)(a2 + ab + b2):
For consecutive integers where a − b = 1:
a3 − (a − 1)3 = 1 + 3a(a − 1).
• (ii) 433 − 423 = 1 + 3 × 43 × 42 = 1 + 3 × 1806 = 1 + 5418 = 5419.
• (i) 673 − 663 = 1 + 3 × 67 × 66 = 1 + 3 × 4422 = 1 + 13266 = 13267. - Overall Comparison:
13267 > 5419 > 133 > 85.
Final Answer: (i) 673 − 663 is the greatest value (equal to 13,267).
Exam Tip: Notice that for consecutive integers n and n − 1, the difference of cubes grows quadratically as ~3n2, whereas the difference of squares grows linearly as ~2n. Hence, for large numbers, difference of cubes will always drastically exceed difference of squares!