Looking for cubes and cube roots class 8? This page gives you the official NCERT Class 8 Mathematics Chapter 6 — Cubes and Cube Roots (English, current edition). The chapter runs about eight PDF pages, the official file is right below, and the explanations here tell you what is inside it before you open it.
The PDF download comes first, then a section-by-section guide to the same chapter. If the chapter did not make sense in class, the explanations below teach each idea in the order the book introduces it.
Download the NCERT Class 8 Cubes and Cube Roots PDF
The file below is the official NCERT Class 8 Mathematics Chapter 6 Cubes and Cube Roots PDF published on ncert.nic.in — it is the full chapter exactly as printed in the current edition, so open it to read the book itself and keep this page beside it for the explanations.
Cubes and Cube Roots Class 8: Chapter at a Glance
Before you open the file, know what is inside it. The table on this page lists how much the chapter holds — its sections, figures, worked examples and exercise questions — so use it to see the chapter’s scale at a glance.
The chapter has two numbered sections, Cubes and Cube Roots. Exercise 6.1 tests number-based skills: recognising perfect cubes and finding the smallest multiplier or divisor. Exercise 6.2 tests cube roots by prime factorisation and true/false reasoning about cubes.
What This Chapter Covers
This is a guided tour. The chapter teaches one operation — cubing — and its reverse, finding the cube root. Each stop below is one idea from the chapter, in the order the book takes them. Section references like 6.1 stay the same across printings; printed page numbers do not.
- Section 6.1 — Introduction: the taxi-number story. Hardy calls 1729 dull; Ramanujan points out it is the smallest number expressible as a sum of two cubes in two different ways.
- Section 6.2 — Cubes: what a cube number is, the cubes from 1 to 20, the unit-digit pattern, and the key discovery that each prime factor of a number appears three times in its cube’s prime factorisation.
- Patterns in cubes: the chapter’s TRY THESE items show \(n^3\) as a sum of consecutive odd numbers, and the shortcut \(n^3 – (n-1)^3 = 1 + 3n(n-1)\).
- Section 6.2.2 — Smallest multiple that is a perfect cube: the fix-it questions. Multiply to complete a missing triplet, or divide to remove a factor that never forms one — with Raj and Anu’s cuboid as the shape version of the same idea.
- Section 6.3 — Cube Roots: cube root as the inverse of cubing, the cube-root symbol \(\sqrt[3]{x}\), and a table matching cubes with their roots.
- Section 6.3.1 — Cube root through prime factorisation: group the factors in triplets, take one factor from each triplet, multiply.
- WHAT HAVE WE DISCUSSED? — the closing summary: four takeaways that this page repeats in plainer words in its revision summary.
Key Concepts: Cubes and Cube Roots, Explained
A cube number is what you get when you multiply a number by itself three times: \(5 \times 5 \times 5 = 125\), so 125 is a cube number. The cube root asks the question backwards — which number, multiplied three times, gives this one? This section runs through the ideas in the same order the chapter does.
Perfect cubes and the cubes from 1 to 20
Cube numbers (also called perfect cubes) are obtained by taking a number three times: \(1^3 = 1\), \(2^3 = 8\), \(3^3 = 27\), \(4^3 = 64\), \(5^3 = 125\). A number like 9 is not a perfect cube because \(2^3 = 8\) and \(3^3 = 27\) — nothing in between gives 9 (Section 6.2).
Two counts are worth remembering. There are only ten perfect cubes from 1 to 1000 — the cubes of 1 to 10. From 1 to 100 there are just four: 1, 8, 27 and 64. Cubes grow fast, so they thin out quickly.
The unit digit rule. The last digit of a cube depends only on the last digit of the base number. The reason the rule works: when two numbers are multiplied, the unit digit of the answer comes only from the unit digits of the factors being multiplied. The map is:
| Last digit of the number | Last digit of its cube |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 7 |
| 4 | 4 |
| 5 | 5 |
| 6 | 6 |
| 7 | 3 |
| 8 | 2 |
| 9 | 9 |
Apply it to the chapter’s TRY THESE numbers: \(8888^3\) ends in 2, because 8 maps to 2, and \(53^3\) ends in 7, because 3 maps to 7. The map also shows that an even base gives an even cube and an odd base gives an odd cube.
Memorise the cubes of 11 to 20 — 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000. The chapter prints them as a table, and knowing them makes the exercise numbers instantly recognisable.
Two patterns that make cubes faster
The chapter stops on two patterns in Section 6.2 that let you write a cube without multiplying three numbers.
Pattern one — cubes as sums of consecutive odd numbers: \(1 = 1^3\), \(3 + 5 = 2^3\), \(7 + 9 + 11 = 3^3\). In general, \(n^3\) is the sum of n consecutive odd numbers starting at \(n^2 – n + 1\).
\[ 8^3 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512 \]
Check the starting number: \(8^2 – 8 + 1 = 57\). This is exactly what the chapter’s TRY THESE asks for when it wants \(6^3\), \(7^3\) and \(8^3\) written as sums of odd numbers.
Pattern two — difference of consecutive cubes: \(2^3 – 1^3 = 1 + 2 \times 1 \times 3\), and in general \(n^3 – (n-1)^3 = 1 + 3n(n-1)\).
\[ 101^3 – 100^3 = 1 + 3 \times 101 \times 100 = 1 + 30300 = 30301 \]
The prime factorisation test for perfect cubes
This is the one reliable test in the chapter. Prime-factorise the number; it is a perfect cube exactly when every prime appears in a group of three (Section 6.2). The book shows \(216 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3\) using the rule \(a^m \times b^m = (a \times b)^m\).
\[ 9261 = 3^3 \times 7^3 = (3 \times 7)^3 = 21^3 \]
So 9261 is a perfect cube — it is 21 cubed. Contrast that with 243, the book’s Example 1: \(243 = 3 \times 3 \times 3 \times 3 \times 3\). Three copies of 3 make one triplet, but two 3s remain, so 243 is not a perfect cube. If any prime is left over, the number fails the test.
Smallest multiplier and smallest divisor
These are fix-it questions (Section 6.2.2): a number is not a perfect cube because some prime appears once or twice instead of three times. Multiply completes a missing triplet; divide removes a straggler prime. The answer the book asks for is the smallest natural number that does the job.
Step 1 — factorise 5400: \(5400 = 2^3 \times 3^3 \times 5^2\).
Step 2 — spot the incomplete triplet: the 2s and 3s are complete; the 5 appears only twice.
Step 3 — multiply by the missing 5: \(5400 \times 5 = 2^3 \times 3^3 \times 5^3 = (2 \times 3 \times 5)^3\).
\[ 5400 \times 5 = 27000 = 30^3 \]
Final answer: the smallest natural number is 5, and 27000 is a perfect cube.
Step 1 — factorise 6912: \(6912 = 2^8 \times 3^3\).
Step 2 — decide divide instead of multiply: the 3s already form a triplet; the eight 2s leave two behind after making two triplets.
Step 3 — divide by the leftover \(2^2 = 4\): \(6912 \div 4 = 2^6 \times 3^3 = (4 \times 3)^3\).
\[ 6912 \div 4 = 1728 = 12^3 \]
Final answer: the smallest natural number is 4, and 1728 is a perfect cube.
The cuboid versions work on volume instead of the number itself. Raj’s cuboid \(15 \times 30 \times 15\) has volume \(6750 = 2 \times 3^3 \times 5^3\) — one 2 short, so four cuboids make a cube. Try a fresh one with sides 5 cm, 5 cm and 10 cm:
Step 1 — volume: \(5 \times 5 \times 10 = 250 = 2 \times 5^3\).
Step 2 — count the missing primes: the 5s form a triplet; the 2 is alone, so we need \(2^2 = 4\) more copies of the cuboid.
Step 3 — total volume: \(4 \times 250 = 1000 = 10^3\).
Final answer: 4 cuboids form a cube of side 10 cm.
Cube roots by prime factorisation
Finding the cube root is the inverse of cubing, exactly as square root is the inverse of squaring (Section 6.3). The symbol \(\sqrt[3]{8} = 2\) is read as “the cube root of 8 is 2”. The method: prime-factorise, group the factors in triplets, take one factor from each triplet, and multiply those chosen factors.
Step 1 — prime factorise 17576: \(17576 = 2^3 \times 13^3\).
Step 2 — take one factor from each triplet: one 2 and one 13.
\[ \sqrt[3]{17576} = 2 \times 13 = 26 \]
Check: \(26 \times 26 \times 26 = 17576\).
Final answer: the cube root of 17576 is 26.
The chapter’s own Examples 6 and 7 do the same for 8000 and 13824 — both factorisations collapse into triplets, giving roots 20 and 24. Small roots come straight from memory: \(\sqrt[3]{27} = 3\), \(\sqrt[3]{64} = 4\).
The 1729 Figure: Hardy–Ramanujan Number Explained
The chapter opens with an illustration of 1729 to make the point that numbers reward attention. The teaching sits in the number, so this is what the figure asks you to notice.
The story (Section 6.1): mathematician G.H. Hardy visited Ramanujan in a taxi numbered 1729 and called it a dull number. Ramanujan answered instantly that it was interesting — it is the smallest number that can be written as a sum of two cubes in two different ways:
\[ 1729 = 1728 + 1 = 12^3 + 1^3 \]
\[ 1729 = 1000 + 729 = 10^3 + 9^3 \]
There are infinitely many such numbers. The chapter gives two more with the pairs in brackets and asks you to check them, so here is the verification:
| Number | First way | Second way |
|---|---|---|
| 1729 | \(12^3 + 1^3\) | \(10^3 + 9^3\) |
| 4104 | \(16^3 + 2^3\) | \(15^3 + 9^3\) |
| 13832 | \(20^3 + 18^3\) | \(24^3 + 2^3\) |
For 4104: \(16^3 + 2^3 = 4096 + 8 = 4104\), and \(15^3 + 9^3 = 3375 + 729 = 4104\). For 13832: \(20^3 + 18^3 = 8000 + 5832 = 13832\), and \(24^3 + 2^3 = 13824 + 8 = 13832\).
One honest note from the book: this feature of 1729 was known more than 300 years before Ramanujan. The point of the story is not that he discovered it — it is that he noticed it, because he experimented with numbers all his life.
Definitions: The Terms Used in This Chapter
A quick glossary for the night before. Every term below is used in the chapter, and most errors in this chapter come from half-remembered definitions.
| Term | Meaning | Where it appears |
|---|---|---|
| Cube number / perfect cube | A natural number obtained by multiplying a number by itself three times, e.g. 1, 8, 27. | Section 6.2 |
| Hardy–Ramanujan number | A number expressible as a sum of two cubes in two different ways; 1729 is the smallest. | Section 6.1 |
| Prime factorisation | Writing a number as a product of prime numbers, e.g. \(216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\). | Section 6.2 |
| Triplet | A group of three equal prime factors. A perfect cube has every prime in a triplet. | Section 6.2 |
| Smallest natural number | The textbook’s phrase for the smallest positive whole number that fixes a non-cube, by multiplication or division. | Section 6.2.2 |
| Cube root | The number whose cube gives the given number; the inverse operation of cubing. | Section 6.3 |
| Cube-root symbol | \(\sqrt[3]{x}\), read as “cube root of x”, e.g. \(\sqrt[3]{27} = 3\). | Section 6.3 |
Common Mistakes in Cubes and Cube Roots
Most errors in this chapter are not arithmetic — they are rules applied to the wrong case. The table names each mistake before you make it, with the chapter’s own evidence and a way to check the fix.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Thinking a number is a perfect cube when only some primes appear three times. | Every prime factor must appear in a triplet. 243 has five 3s — after grouping one triplet, two 3s remain, so 243 is not a perfect cube (Example 1, Section 6.2). | Factorise completely and count each prime; nothing may be left outside a triplet. |
| Using the multiply fix where the question asks for the divide fix. | Multiply completes a missing triplet; divide removes a prime that never reaches three copies (Examples 2–5, Section 6.2.2). | After your fix, factorise again; every prime must sit inside a triplet. |
| Assuming a cube ending in 8 has a base ending in 8. | The unit digits of 2 and 8 swap when cubed: a cube ending in 8 must have a base ending in 2 (TRY THESE, Section 6.2). | Apply the unit-digit map before multiplying: 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9, 0→0. |
| Believing \(m^2 \lt m^3\) for every integer m. | False for 0, 1 and every negative integer: \(0^2 = 0^3\), \(1^2 = 1^3\), and \((-2)^2 = 4 \gt (-2)^3 = -8\) (THINK, DISCUSS AND WRITE, Section 6.3). | Test m = 0, m = 1 and a negative number before generalising. |
| Thinking a perfect cube can end with exactly two zeros. | \(10 = 2 \times 5\), and both primes must appear in triplets, so trailing zeros come in threes — never two (Exercise 6.2, Q2(ii)). | Count the 2s and 5s in the prime factorisation; each needs a triplet. |
| Reasoning “square ends in 5, so cube ends in 25”. | False — \(15^2 = 225\) but \(15^3 = 3375\), which ends in 75 (Exercise 6.2, Q2(iii)). | Cube the number once: \(15 \times 15 \times 15\). |
And for Exercise 6.2 Q2, write the reason, not just true or false. Every statement in that set is settled by one of the rules above — the unit-digit map, the triplet rule, or a quick cube of a small number.
Exam Notes: The Skills This Chapter Builds
No chapter guarantees any particular question, but this one builds exactly two skills, and the exercises are arranged around them. Practising the exercises in order is the honest preparation.
- Exercise 6.1 — number-based. Q1: which numbers are not perfect cubes (216, 128, 1000, 100, 46656). Q2: smallest number to multiply. Q3: smallest number to divide. Q4: the cuboid-to-cube question.
- Exercise 6.2 — cube roots and reasoning. Q1: ten cube roots by prime factorisation, from 64 to 91125. Q2: seven true/false statements about cubes.
A full-marks answer to a multiplier or divisor question shows the factorisation with triplets circled, names the missing or leftover prime, and states the smallest natural number as the final line. A full-marks cube-root answer groups the factorisation into triplets, takes one factor from each, and writes the product.
The step format worth memorising: factorise → circle the triplets → read the missing or leftover primes. It works for every multiplier, divisor and cube-root question in the chapter.
Speed comes from two memorised tools: the cubes of 1 to 20 and the unit-digit map. Together they let you check any answer in this chapter in seconds — the cube of a two-digit number can be cross-checked by its size and its last digit.
One caveat: textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus before deciding what to study.
Revision Summary: Cubes and Cube Roots in Five Lines
Built from the chapter’s own closing summary, WHAT HAVE WE DISCUSSED? — five lines that carry the whole chapter.
- Hardy–Ramanujan numbers — 1729, 4104, 13832 — can be written as a sum of two cubes in two different ways; 1729 is the smallest.
- A cube number comes from multiplying a number by itself three times: 1, 8, 27 are cube numbers.
- A number is a perfect cube exactly when each prime factor appears three times in its prime factorisation.
- The symbol \(\sqrt[3]{x}\) denotes the cube root, the inverse of cubing: \(\sqrt[3]{27} = 3\).
- The exercises need two working methods: check a cube by factorising into triplets; find a cube root by grouping, then taking one factor from each triplet.
Related Resources
This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.
If you are revising the whole book, the pages next to this chapter are the ones to open next.
- Class 8 Mathematics book page — all chapters of this book, including Chapter 6.
- Class 8 hub — every NCERT Class 8 subject listed.
- Previous chapter: Squares and Square Roots — the companion chapter on the square operation and its inverse.
- Next chapter: Comparing Quantities — ratios, percentages and profit and loss.
- Squares and Square Roots revision notes — for the chapter that leads straight into this one.
- Class 8 mathematics notes and the Class 8 notes hub hold the test-style revision pages; the CBSE notes home lists every class.
- Comparing Quantities (proportional reasoning) notes — the next chapter’s notes page.
For the whole book rather than one chapter, the official NCERT Class 8 Mathematics textbook page on ncert.nic.in lists every chapter PDF in the book.
Sources and Data Verification
- The chapter contents and figure on this page describe the NCERT Class 8 Mathematics textbook, Chapter 6, Cubes and Cube Roots, English, current edition as published on ncert.nic.in.
- This page covers that one chapter only — not the whole book, and not the CBSE syllabus.
- The page is maintained for the current academic session using the NCERT information available to us. Textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus.
- NCERT settles the textbook — its chapters, editions and PDFs. CBSE settles curriculum, syllabus and examinations. Section references (6.1 to 6.3.1) are used rather than printed page numbers because page numbering varies between printings.
| What the chapter holds | Count | Where it is used |
|---|---|---|
| Printed pages | 8 | |
| Sections in the chapter | 6 | |
| Figures with NCERT captions | 1 | |
| Tables | 6 | |
| Worked examples | 1 | solved step by step in our NCERT Solutions |
| Exercise questions | 6 | answered in our NCERT Solutions |
| Official NCERT PDF | Download the chapter PDF | the chapter exactly as NCERT publishes it |
Cubes and Cube Roots Class 8: FAQs
How do you find the smallest number by which a non-cube must be multiplied to obtain a perfect cube?
Prime-factorise the number and look for primes that do not appear in a group of three, then multiply by each missing prime once. Example: \(5400 = 2^3 \times 3^3 \times 5^2\); the 5 needs one more copy, so the smallest multiplier is 5, giving \(27000 = 30^3\).
How do you find the cube root of a large number using prime factorisation?
Factorise the number, group the factors in triplets, take one factor from each triplet, and multiply them. Example: \(17576 = 2^3 \times 13^3\), so \(\sqrt[3]{17576} = 2 \times 13 = 26\).
Why is 1729 called the Hardy–Ramanujan number?
Because the taxi number Hardy called dull is the smallest number expressible as a sum of two cubes in two different ways: \(1729 = 12^3 + 1^3 = 10^3 + 9^3\). Ramanujan spotted this instantly; the chapter adds that the property was known more than 300 years before him.
How can you predict the last digit of a cube without calculating the whole cube?
Only the last digit of the base matters, because the unit digit of a product depends only on the unit digits of the factors. Use the map \(1 \rightarrow 1\), \(2 \rightarrow 8\), \(3 \rightarrow 7\), \(4 \rightarrow 4\), \(5 \rightarrow 5\), \(6 \rightarrow 6\), \(7 \rightarrow 3\), \(8 \rightarrow 2\), \(9 \rightarrow 9\), \(0 \rightarrow 0\). So \(8888^3\) ends in 2 and \(53^3\) ends in 7 without multiplying anything out.
Why does a perfect cube never end with exactly two zeros?
Because a zero at the end comes from a factor of \(10 = 2 \times 5\), and in a perfect cube every prime factor must appear three times. So the 2s and 5s must both come in triplets, which forces the number of trailing zeros to be a multiple of three — three, six, nine, never two.
The chapter tests exactly this in Exercise 6.2 Q2(ii), with \(1000 = 10^3\) ending in three zeros.
Reference: NCERT Class 8 Mathematics textbook, chapter 6, official edition on ncert.nic.in.
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