1. Let’s compare using identities: a³ – b³ = (a–b)(a² + ab + b²) 67³ – 66³ = 1(4489 + 4422 + 4356) = 13267 43³ – 42³ = 1(1849 + 1806 + 1764) = 5419 a² – b² = (a–b)(a + b) 67² – 66² = 1(133) = 133 43² – 42² = 1(85) = 85 So, the greatest is 67³ – 66³   For more NCERT Solutions for Class 8 MathematicRead more

    Let’s compare using identities:
    a³ – b³ = (a–b)(a² + ab + b²)
    67³ – 66³ = 1(4489 + 4422 + 4356) = 13267
    43³ – 42³ = 1(1849 + 1806 + 1764) = 5419
    a² – b² = (a–b)(a + b)
    67² – 66² = 1(133) = 133
    43² – 42² = 1(85) = 85
    So, the greatest is 67³ – 66³

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 95
  2. We can guess these cube roots: 1331 = 11³ (last digit is 1) 4913 = 17³ 12167 = 23³ 32768 = 32³ By knowing cube values or estimating from digit patterns, we match these cubes to their roots. For instance, 32³ = 32768, so cube root is 32. This helps in quick identification without full factorisation.Read more

    We can guess these cube roots:
    1331 = 11³ (last digit is 1)
    4913 = 17³
    12167 = 23³
    32768 = 32³
    By knowing cube values or estimating from digit patterns, we match these cubes to their roots. For instance, 32³ = 32768, so cube root is 32. This helps in quick identification without full factorisation.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 99
  3. (i) False – Odd³ = Odd (e.g., 3³ = 27) (ii) False – 2³ = 8 ends in 8 (iii) True – 4³ = 64 (2-digit → 2-digit), 5³ = 125 (2-digit → 3-digit) (iv) True – 99³ = 970299 (2-digit cube has 6 digits) (v) False – Only cube numbers that are also squares have odd number of factors. Others have even.   FoRead more

    (i) False – Odd³ = Odd (e.g., 3³ = 27)
    (ii) False – 2³ = 8 ends in 8
    (iii) True – 4³ = 64 (2-digit → 2-digit), 5³ = 125 (2-digit → 3-digit)
    (iv) True – 99³ = 970299 (2-digit cube has 6 digits)
    (v) False – Only cube numbers that are also squares have odd number of factors. Others have even.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 102
  4. 1323 = 3³ × 7². For it to be a perfect cube, each prime must occur in multiples of 3. We already have 7², so we need one more 7 to make it 7³. Therefore, multiply 1323 by 7 to get 3³ × 7³ = (3×7)³ = 21³ = 9261 Answer: Multiply by 7 and cube root of the result is 21.   For more NCERT Solutions fRead more

    1323 = 3³ × 7². For it to be a perfect cube, each prime must occur in multiples of 3.
    We already have 7², so we need one more 7 to make it 7³.
    Therefore, multiply 1323 by 7 to get 3³ × 7³ = (3×7)³ = 21³ = 9261
    Answer: Multiply by 7 and cube root of the result is 21.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 27
  5. To find cube roots: 27000 = 3 × 3 × 3 × 10 × 10 × 10 = (3×10)³ = 30³ ⇒ √³27000 = 30 10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2×11)³ = 22³ ⇒ √³10648 = 22 So, cube roots are 30 and 22 respectively. Both numbers are perfect cubes.   For more NCERT Solutions for Class 8 Mathematics Ganita Prakash ChapteRead more

    To find cube roots:
    27000 = 3 × 3 × 3 × 10 × 10 × 10 = (3×10)³ = 30³ ⇒ √³27000 = 30
    10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2×11)³ = 22³ ⇒ √³10648 = 22
    So, cube roots are 30 and 22 respectively. Both numbers are perfect cubes.

     

    For more NCERT Solutions for Class 8 Mathematics Ganita Prakash Chapter 1 A Square and A Cube Extra Questions & Answer:

    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/ganita-prakash-chapter-1/

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    • 100