Figure it Out — Cubes and Cube Roots — A Square and A Cube

1. 12 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.3), Q1

   Find the cube roots of 27000 and 10648.

   Hint. Try to spot each number as a simple multiple of a cube you already know, or prime-factorise directly.

   Step 1 — Factorise 27000. 27000 = 27 × 1000 = 3³ × 10³ = (3 × 10)³ = 30³.

   Step 2 — Factorise 10648. 10648 = 2³ × 1331 = 2³ × 11³ = (2 × 11)³ = 22³.

   ✦ Answer: ∛27000 = 30 and ∛10648 = 22.

   Where students slip. Trying long division-style estimation for these — both numbers factor cleanly into a product of small cubes once split into recognisable pieces (1000 and 1331).

2. 22 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.3), Q2

   What number will you multiply by 1323 to make it a cube number?

   Hint. Prime-factorise 1323 and see which prime's exponent isn't already a multiple of 3.

   Step 1 — Prime factorise 1323. 1323 = 3 × 441 = 3 × 3² × 7² = 3³ × 7².

   Step 2 — Check the exponents. 3³ is already a complete triplet; 7² is missing one more 7 to complete its triplet.

   Step 3 — Multiply to complete it. 1323 × 7 = 9261 = 3³ × 7³ = (3 × 7)³ = 21³, since every prime's exponent is now a multiple of 3.

   ✦ Answer: Multiply by 7 (giving 9261 = 21³).

   Where students slip. Multiplying by the whole number 1323 or by 3 — only the prime whose exponent isn't yet a multiple of 3 (here, the 7s) needs completing.

3. 33 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.3), Q3

   State true or false, explaining your reasoning: (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.

   Hint. For (iii) and (iv), check the cubes of the smallest and largest 2-digit numbers to bound how many digits a 2-digit number's cube can have.

   Step 1 — Check (i). Odd × odd × odd is always odd (e.g. 3³ = 27), so the cube of an odd number is never even — False.

   Step 2 — Check (ii). 2³ = 8 and 12³ = 1728 both end in 8, so perfect cubes ending in 8 do exist — False.

   Step 3 — Check (iii) and (iv) using the smallest and largest 2-digit numbers. 10³ = 1000 (4 digits) is already the smallest possible 2-digit cube, so no 2-digit number's cube has only 3 digits — (iii) is False. 99³ = 970299 (6 digits) is the largest, so no 2-digit number's cube reaches 7 digits — (iv) is False.

   Step 4 — Check (v). An odd number of factors is a property of perfect squares, not cubes in general — e.g. 8 = 2³ has factors 1, 2, 4, 8, which is 4 (even) factors — False.

   ✦ Answer: All five statements are False.

   Where students slip. Assuming a 2-digit number's cube could be anywhere from 3 to 7 digits — checking the actual smallest (10³ = 1000) and largest (99³ = 970299) cases pins the range down to exactly 4-6 digits.

4. 42 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.3), Q4

   You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

   Hint. Use the units digit to guess the cube root's units digit, and the overall size of the number to guess its tens digit.

   Step 1 — Estimate 1331's cube root. It's between 10³ = 1000 and 20³ = 8000, and its units digit (1) means the root ends in 1 — so 11 is a good guess; checking, 11³ = 1331 ✓.

   Step 2 — Estimate 4913's cube root. Between 10³ = 1000 and 20³ = 8000; units digit 3 suggests a root ending in 7 (since 7³ = 343 ends in 3) — 17³ = 4913 ✓.

   Step 3 — Estimate 12167's cube root. Between 20³ = 8000 and 30³ = 27000; units digit 7 suggests a root ending in 3 (since 3³ = 27 ends in 7) — 23³ = 12167 ✓.

   Step 4 — Estimate 32768's cube root. Between 30³ = 27000 and 40³ = 64000; units digit 8 suggests a root ending in 2 (since 2³ = 8) — 32³ = 32768 ✓.

   ✦ Answer: ∛1331 = 11, ∛4913 = 17, ∛12167 = 23, ∛32768 = 32.

   Where students slip. Trying full prime factorisation for each — the question specifically asks for a guess using units-digit and size clues, which is faster once the units-digit-to-units-digit correspondence for cubes is known.

5. 53 marksGanita Prakash Cl-8 Part 1, Figure it Out (§1.3), Q5

   Which of the following is the greatest? Explain your reasoning. (i) 67³ − 66³ (ii) 43³ − 42³ (iii) 67² − 66² (iv) 43² − 42²

   Hint. Use the identity a³ − b³ = (a − b)(a² + ab + b²) and a² − b² = (a − b)(a + b) instead of computing each cube or square directly.

   Step 1 — Expand the cube differences. 67³ − 66³ = (67 − 66)(67² + 67×66 + 66²) = 1 × (4489 + 4422 + 4356) = 13267. 43³ − 42³ = (43 − 42)(43² + 43×42 + 42²) = 1 × (1849 + 1806 + 1764) = 5419.

   Step 2 — Expand the square differences. 67² − 66² = (67 − 66)(67 + 66) = 133. 43² − 42² = (43 − 42)(43 + 42) = 85.

   Step 3 — Compare all four values. 13267 > 5419 > 133 > 85.

   ✦ Answer: (i) 67³ − 66³ = 13267 is the greatest, since consecutive-cube differences grow far faster than consecutive-square differences, and 67 gives a larger difference than 43 in both cases.

   Where students slip. Assuming the two square differences (which look 'simpler') might compete with the cube differences — cube differences between consecutive numbers are on a completely different scale from square differences, since they involve a sum of three squared-ish terms rather than a simple sum.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 1, Reprint 2026-27 (hegp101.pdf). This chapter's questions are scattered as 'Math Talk'/'Try This' prompts through the running text, plus two formal 'Figure it Out' exercise blocks. The book supplies its own answer key at the end of the chapter for every question that has one definite answer — that answer key is the ground truth this file is checked against. One question (Figure it Out — Squares, Q9, the tiny-squares picture) has no answer in the book's own key; it was solved here by rendering the actual figure and counting directly (9×8 = 72 tiles of a 5×5 grid each = 1800 tiny squares).. Questions are referenced from the NCERT textbook for identification.