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NCERT Solutions Ganita Prakash (Part 1) Chapter 1 Cube Roots; Successive Differences — In-text Questions

Book page 15 Updated on2026-09-05

Q1.
Find the cube roots of these numbers: (i) ∛64 = (ii) ∛512 = (iii) ∛729 =
Answer

Factorise each into triplets.

(i) 64 = 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2)3 = 43 → ∛64 = 4
(ii) 512 = 29 = (23)3 = 83 → ∛512 = 8
(iii) 729 = 36 = (32)3 = 93 → ∛729 = 9
Why exponents make this quick: Each of these is a power of a single prime, so taking the cube root just divides the exponent by 3: 26 → 22, 29 → 23, 36 → 32.
Did you know? 64 is both a perfect square (82) and a perfect cube (43), because 64 = 26 and 6 is divisible by both 2 and 3. The next such number is 729 = 36 = 272 = 93.
Q2.
Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?
Answer

Start from the cubes and keep differencing.

Perfect cubes182764125216343
Level 1719376191127
Level 21218243036
Level 36666

The differences become constant at level 3, and the constant is 6.

Why level 3, and why 6: For squares, one round of differencing turns n2 into 2n + 1 (degree 1) and a second round gives the constant 2. For cubes,
level 1: (n+1)3 – n3 = 3n2 + 3n + 1, a degree-2 expression;
level 2: differencing that gives 6n + 6, a degree-1 expression;
level 3: differencing again gives the constant 6.
Each round of differencing lowers the degree by one, so a third power needs three rounds. The constant left behind is 3 × 2 × 1 = 6, just as squares leave 2 × 1 = 2.
Tip: Predict before you compute — fourth powers would flatten out at level 4, with the constant 4 × 3 × 2 × 1 = 24. Try it on 1, 16, 81, 256, 625, 1296.
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