In-text Questions — Perfect Cubes and Cube Roots — A Square and A Cube

1. 12 marksGanita Prakash Cl-8 Part 1, Math Talk in §1.2

   Can a cube end with exactly two zeroes (00)? Explain.

   Hint. Recall that squares can only ever have an even number of trailing zeros — is there a similar rule for cubes?

   Step 1 — Recall where trailing zeros come from. Trailing zeros come from factors of 10 = 2 × 5 in a number.

   Step 2 — Apply this to a perfect cube. Since a perfect cube is n × n × n, every prime factor's exponent (including those of 2 and 5) must be a multiple of 3 — so the count of trailing zeros must also be a multiple of 3 (0, 3, 6, 9, ...).

   Step 3 — Check whether 2 fits. 2 is not a multiple of 3.

   ✦ Answer: No — a perfect cube's trailing-zero count must always be a multiple of 3, so exactly two trailing zeros is impossible.

   Where students slip. Assuming cubes follow the same 'even number of zeros' rule as squares — cubes instead require the zero-count to be a multiple of 3, not just an even number.

2. 22 marksGanita Prakash Cl-8 Part 1, Try This in §1.2, on consecutive-odd-number sums for cubes

   Later in the consecutive-odd-numbers-sum-to-cubes series, we reach 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. Can you tell what this sum is without doing the calculation?

   Hint. Count how many terms are in this sum, and recall which cube that number of terms corresponds to.

   Step 1 — Recall the pattern. The sum of the first n terms of this consecutive-odd-number series (starting fresh after each previous cube) equals n³.

   Step 2 — Count the terms given. 91, 93, 95, 97, 99, 101, 103, 105, 107, 109 — that's 10 terms.

   Step 3 — Apply the pattern. 10 terms correspond to 10³, since this run is exactly the 10th group in the series (the terms just keep going up from where the 9th group left off).

   ✦ Answer: The sum is 10³ = 1000.

   Where students slip. Adding up all ten numbers directly — counting how many terms are in the run and matching that count to its cube avoids the arithmetic entirely.

3. 32 marksGanita Prakash Cl-8 Part 1, prompt after the cube-root-via-prime-factorisation examples

   Find the cube roots of these numbers: (i) 64 (ii) 512 (iii) 729

   Hint. Prime-factorise each number and group the factors into triplets.

   Step 1 — Factorise 64. 64 = 2⁶ = (2²)³, so its cube root is 2² = 4.

   Step 2 — Factorise 512. 512 = 2⁹ = (2³)³, so its cube root is 2³ = 8.

   Step 3 — Factorise 729. 729 = 3⁶ = (3²)³, so its cube root is 3² = 9.

   ✦ Answer: ∛64 = 4, ∛512 = 8, ∛729 = 9.

   Where students slip. Grouping the prime factors into pairs (as for square roots) instead of triplets — cube roots need the factors split into groups of three, not two.

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.