Cubes and Cube Roots

In 1919, a mathematician lying sick in a nursing home was visited by his friend, who mentioned that the number of his taxi, 1729, seemed "rather a dull number." The patient, Srinivasa Ramanujan, disagreed instantly: 1729 is the smallest number expressible as the sum of two cubes in two different ways, $1729 = 1^3 + 12^3 = 9^3 + 10^3$ . That flash of insight is only possible because Ramanujan had spent a lifetime noticing patterns in cubes — and this chapter hands you the same toolkit on a Class 8 scale. A cube number is what you get when a number is multiplied by itself three times: $1, 8, 27, 64, 125, \ldots$ are the cubes of $1, 2, 3, 4, 5$ . Geometrically, a perfect cube is the volume of a cube-shaped solid with a whole-number side. This chapter asks two natural questions about such numbers. First, given a number, how do you recognise whether it is a perfect cube, and if it is not, what is the smallest number you must multiply or divide by to make it one? The key tool is prime factorisation: in a perfect cube, every prime factor occurs in a group of exactly three. Second, given a perfect cube, how do you find the number whose cube it is — the cube root, written with the symbol $\sqrt[3]{\ }$ ? This is the reverse of cubing, just as square roots reverse squaring. Along the way you will meet beautiful patterns: cubes built from sums of consecutive odd numbers, a rule linking consecutive cubes, and quick ways to spot the units digit of a cube. By the end you will be able to test any number for being a perfect cube, adjust it into one, and extract cube roots by prime factorisation — skills that quietly return in mensuration, algebra and later in exponents and surds.

Key takeaways

A perfect cube (cube number) is obtained by multiplying a number by itself three times: $n^3 = n \times n \times n$ .
The cubes of $1$ to $10$ are $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000$ — worth memorising.
Test for a perfect cube: write the prime factorisation; the number is a perfect cube exactly when every prime factor appears in a group of three.
Cubes of even numbers are even; cubes of odd numbers are odd — the parity of a number matches the parity of its cube.
Consecutive odd numbers sum to consecutive cubes: $1 = 1^3$ , $3+5 = 2^3$ , $7+9+11 = 3^3$ , and so on — the $n$ -th cube needs $n$ consecutive odd numbers.
To fix a non-cube number, find the prime(s) whose count is not a multiple of three, then multiply or divide by just enough copies of that prime to complete the triple.
Cube root is the inverse of cubing: $\sqrt[3]{n^3} = n$ . Found by grouping prime factors in triples and taking one factor from each triple.
$1729$ , the Hardy-Ramanujan Number, is the smallest number expressible as a sum of two cubes in two different ways: $1729 = 1^3 + 12^3 = 9^3 + 10^3$ .
Key idea: a number is a perfect cube exactly when its prime factorisation splits into complete triples, and the cube root is recovered by taking one factor from each triple.

The core insight

A number is a perfect cube exactly when every prime factor in its factorisation occurs in a group of three, and the cube root is found by taking one factor from each such triple.

What this chapter covers

The chapter builds in three connected stages. Cubes opens by defining a cube number geometrically and arithmetically, tabulates the cubes of $1$ to $20$ , and uncovers two number patterns — cubes as sums of consecutive odd numbers, and a rule for the difference of consecutive cubes. Cubes and prime factorisation turns this into a working test: a number is a perfect cube exactly when its prime factorisation groups completely into triples; you use this to check whether a given number is a perfect cube and, if not, to find the smallest multiplier or divisor that fixes it. Cube roots then reverses the whole process: given a perfect cube, prime factorisation recovers the cube root by pulling one factor from each triple. Throughout, keep the connection between the geometric cube (a solid of equal sides) and the arithmetic cube ( $n^3$ ) in view — it is what makes the "smallest cuboid multiplier" problems (like Raj's plasticine cuboid) make sense.

The core ideas

Cube numbers: definition and the first ten

A cube or perfect cube is the result of multiplying a natural number by itself, three times over. If $n$ is a natural number, its cube is written $n^3$ and read " $n$ cubed":

n^3 = n \times n \times n.

where: $n$ is any natural number, and $n^3$ is called a perfect cube or cube number. Geometrically this matches a solid cube of side $n$ units built from unit cubes — a cube of side $2$ needs $2^3 = 8$ unit cubes, a cube of side $3$ needs $3^3 = 27$ unit cubes, and so on. NCERT's Table 1 lists the cubes of $1$ to $10$ :

1^3=1,\ 2^3=8,\ 3^3=27,\ 4^3=64,\ 5^3=125,\ 6^3=216,\ 7^3=343,\ 8^3=512,\ 9^3=729,\ 10^3=1000.

Only these ten perfect cubes lie between $1$ and $1000$ — perfect cubes thin out fast because cubing grows so much faster than squaring. A useful check for whether a number could be a cube is to notice that cubes of even numbers are always even (e.g. $6^3 = 216$ ) and cubes of odd numbers are always odd (e.g. $7^3 = 343$ ): the parity of $n$ and $n^3$ always agree.

The odd-number pattern

Cubes hide a surprisingly clean pattern when you write them as sums of consecutive odd numbers, starting the count where the previous cube's odd numbers left off:

1 = 1^3,\quad 3+5 = 2^3,\quad 7+9+11 = 3^3,\quad 13+15+17+19 = 4^3,\quad 21+23+25+27+29 = 5^3.

where: the $n$ -th cube is the sum of exactly $n$ consecutive odd numbers. Notice how many odd numbers are used grows by one each time: one for $1^3$ , two for $2^3$ , three for $3^3$ . This gives a quick way to check a guessed cube — if you can split it into the right count of consecutive odd numbers starting where the pattern demands, it confirms the cube.

Prime factorisation and the triple test

The single most useful idea in this chapter is a test for perfect cubes using prime factorisation. Consider the prime factorisation of a cube like $12^3$ :

12^3 = (2\times2\times3)^3 = 2\times2\times2\times2\times2\times2\times3\times3\times3 = 2^6 \times 3^3.

where: every prime factor of $12$ appears exactly three times as often in $12^3$ — this follows from the exponent rule $a^m \times b^m = (a\times b)^m$ applied backwards. This observation reverses into a test: a number is a perfect cube if and only if, in its prime factorisation, every distinct prime factor appears in a group (multiple) of three. If even one prime is short of a full triple, the number is not a perfect cube — but multiplying or dividing by the right number of copies of that prime can fix it.

Figure

Unit cubes stacked to build cubes of side 1 2 3 and 4 showing the volumes 1 8 27 and 64

Cube roots as the inverse of cubing

Finding a cube root undoes cubing, exactly as finding a square root undoes squaring. If $n^3 = m$ , then the cube root of $m$ is $n$ , written using the radical symbol $\sqrt[3]{\ }$ :

\sqrt[3]{n^3} = n.

where: $\sqrt[3]{\ }$ denotes "cube root of," so $\sqrt[3]{8} = \sqrt[3]{2^3} = 2$ and $\sqrt[3]{125} = \sqrt[3]{5^3} = 5$ . To find the cube root of a perfect cube by the prime factorisation method: write the number as a product of primes, group the identical primes into triples, and take exactly one factor from each triple; multiplying those chosen factors together gives the cube root. For example, $3375 = 3\times3\times3\times5\times5\times5 = 3^3\times5^3 = (3\times5)^3$ , so $\sqrt[3]{3375} = 3\times5 = 15$ .

Figure

Number line marking the perfect cubes 1 8 27 64 125 with unequal spacing showing how fast cubes grow

Testing and repairing perfect cubes

Derivation AThe triple-grouping test for a perfect cube

The goal of this method is to decide, for any given natural number, whether it is a perfect cube, and if it is not, to find the smallest natural number by which it must be multiplied (or divided) so that the result becomes a perfect cube. The method rests entirely on prime factorisation and the fact, seen above, that in the prime factorisation of a perfect cube every distinct prime occurs exactly a multiple of three times, because cubing a number cubes each of its prime factors. We illustrate the multiplying case with $392$ and the dividing case with $53240$ , matching NCERT Examples 2 and 3.

Step 1. Write the prime factorisation of the given number using repeated division by primes. For $392$ : $392 = 2\times2\times2\times7\times7 = 2^3 \times 7^2$ . For $53240$ : $53240 = 2\times2\times2\times11\times11\times11\times5 = 2^3\times11^3\times5^1$ .

Step 2. Group each distinct prime's copies into triples and identify any prime left short of a full triple. In $392 = 2^3\times7^2$ , the prime $2$ already forms a complete triple, but $7$ appears only twice — one short of a triple. In $53240 = 2^3\times11^3\times5^1$ , both $2$ and $11$ form complete triples, but $5$ appears only once.

Step 3. Decide the smallest correcting factor. To multiply $392$ into a cube, supply exactly the missing copies of the short prime: $7$ needs one more $7$ to complete its triple, so multiply by $7$ . Check: $392\times7 = 2^3\times7^3 = (2\times7)^3 = 14^3 = 2744$ , a perfect cube. To divide $53240* into a cube, instead remove the prime that cannot form a triple entirely: since$ 5 $appears only once (not enough to complete or discard part of a triple), divide it out completely by dividing by$ 5 $. Check:$ 53240 \div 5 = 2^3\times11^3 = (2\times11)^3 = 22^3 = 10648$, a perfect cube.

Step 4. State the general rule extracted from these two cases. If a prime's exponent leaves remainder $1$ after dividing by $3$ , then to make a cube by multiplication you need two more copies of that prime (or, equivalently, to make a cube by division you remove that one leftover copy, provided it is the only occurrence). If the remainder is $2$ , you need one more copy to multiply up to the next triple, or you must divide out both leftover copies. The smallest multiplier or divisor is always built only from the primes whose counts are not already multiples of three — primes already in complete triples are left untouched. Answer pattern: $392$ is not a perfect cube; multiplying by $7$ gives the perfect cube $2744$ . $53240$ is not a perfect cube; dividing by $5$ gives the perfect cube $10648$ .

Go deeper · Derivation B · 1 minCube root of a perfect cube by prime factorisation

Derivation BCube root of a perfect cube by prime factorisation

The goal here is to extract the cube root of a large perfect cube directly from its prime factorisation, without guessing or long searches, by exploiting the fact that a perfect cube's prime factorisation splits exactly into triples. This is the inverse process of the multiplying/dividing method above: instead of repairing a broken factorisation, we take a complete one and collapse each triple down to a single factor. We work through NCERT's own $13824$ , matching Example 7.

Step 1. Find the complete prime factorisation of the number by repeated division. For $13824$ , repeated division by $2$ and then $3$ gives $13824 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$ , that is, nine $2$ 's and three $3$ 's.

Step 2. Confirm the number is indeed a perfect cube by checking that every prime's exponent is a multiple of three. Here $2$ occurs $9 = 3\times3$ times and $3$ occurs $3 = 3\times1$ times, so both exponents are multiples of three, and $13824$ is confirmed a perfect cube.

Step 3. Group each prime's occurrences into triples and rewrite the factorisation as a product of cubes. The nine $2$ 's split into three triples, each triple being $2^3$ , so $2^9 = 2^3\times2^3\times2^3$ ; the three $3$ 's form one triple, $3^3$ . So $13824 = 2^3\times2^3\times2^3\times3^3$ .

Step 4. Take exactly one factor from each triple and multiply these chosen factors together; this product is the cube root. From the three triples of $2$ take one $2$ each, contributing $2\times2\times2 = 8$ ; from the one triple of $3$ take one $3$ . Multiplying, $\sqrt[3]{13824} = 2\times2\times2\times3 = 24$ . Check by cubing back: $24^3 = 24\times24\times24 = 13824$ , confirming the answer. Answer pattern: $13824 = 2^9\times3^3$ , grouped as $2^3\times2^3\times2^3\times3^3$ , so $\sqrt[3]{13824} = 2\times2\times2\times3 = 24$ .

Worked examples

Board level - checking a perfect cube and fixing it by multiplication (NCERT Example 1 style)

Is $243$ a perfect cube? If not, find the smallest number by which it must be multiplied to make it a perfect cube.

Step 1. Factorise $243$ into primes: $243 = 3\times3\times3\times3\times3 = 3^5$ .

Step 2. Group the five $3$ 's into triples: one complete triple $3^3$ is formed, leaving $3\times3$ ungrouped.

Step 3. Since two extra $3$ 's remain (not a full triple), $243$ is not a perfect cube. To complete the second triple we need one more $3$ .

Step 4. Multiply: $243\times3 = 3^6 = (3^2)^3 = 9^3 = 729$ , which is a perfect cube. Answer: $243$ is not a perfect cube; the smallest multiplier is $3$ , giving the perfect cube $729$ .

Board level - cube root of 74088 by prime factorisation (NCERT style)

Find $\sqrt[3]{74088}$ using the prime factorisation method.

Step 1. Factorise $74088$ by repeated division: $74088 = 2\times2\times2\times3\times3\times3\times7\times7\times7$ .

Step 2. Group into triples: $74088 = 2^3\times3^3\times7^3 = (2\times3\times7)^3$ .

Step 3. Take one factor from each triple and multiply: $2\times3\times7 = 42$ .

Step 4. Check by cubing: $42^3 = 42\times42\times42 = 74088$ , confirming the result. Answer: $\sqrt[3]{74088} = 42$ .

Higher-order thinking (HOTS / olympiad-style) - the smallest cuboid multiplier and a Hardy-Ramanujan check

Parikshit makes a cuboid of plasticine with sides $5\ \text{cm}$ , $2\ \text{cm}$ , $5\ \text{cm}$ . (a) How many such cuboids does he need, laid together, to form a perfect cube? (b) Separately, verify that $1729$ , the Hardy-Ramanujan Number, equals $1^3+12^3$ as well as $9^3+10^3$ .

Step 1. Find the volume of one cuboid and its prime factorisation: volume $= 5\times2\times5 = 50 = 2\times5\times5 = 2^1\times5^2$ .

Step 2. Identify the shortfall in each prime's count to reach a triple: the prime $2$ has exponent $1$ (needs two more copies of $2$ ), and the prime $5$ has exponent $2$ (needs one more copy of $5$ ). Stacking $k$ identical cuboids multiplies the volume by $k$ , so we need $k = 2\times2\times5 = 20$ to supply the missing $2^2\times5^1$ ; combined volume becomes $2^3\times5^3 = (2\times5)^3 = 1000 = 10^3$ .

Step 3. For the Hardy-Ramanujan check, compute both sums directly. $1^3+12^3 = 1 + 1728 = 1729$ . Also $9^3+10^3 = 729 + 1000 = 1729$ . Both equal $1729$ , confirming it is expressible as a sum of two cubes in two genuinely different ways, which is exactly why Ramanujan called it interesting rather than dull.

Step 4. State both final answers together: the cuboid count comes from completing every prime's triple, and the Hardy-Ramanujan identity comes from direct cube arithmetic — two different flavours of "look at the cube structure of a number." Answer: (a) $20$ cuboids are needed to form a cube of side $10$ cm; (b) $1729 = 1^3+12^3 = 9^3+10^3$ , confirmed. Timing note: about three to four minutes total. For part (a), always factorise the volume, not the individual sides, and read off the shortfall prime by prime — a common slip is to try to "cube each side separately," which does not track how many identical cuboids are needed. For part (b), the trap is arithmetic slips in cubing two-digit numbers; keep $12^3 = 1728$ and $10^3=1000$ memorised as anchors since they recur throughout this chapter.

Common traps

A frequent mistake is checking only whether a number is even or divisible by a small prime once, rather than fully factorising and checking that every prime's exponent is a multiple of three — a number like $500 = 2^2\times5^3$ fails only because of the $2$ 's, and a partial check can miss this. Another trap is confusing the multiply-to-fix and divide-to-fix procedures: when a prime's leftover count after removing complete triples is $1$ , dividing out just that one leftover copy works, but multiplying to fix it requires two more copies, not one — students often multiply by the same number they should have divided by. In the odd-number-sum pattern, a common slip is starting the next block of odd numbers from the wrong place; each new cube's block continues immediately after the previous block ended, it does not restart from $1$ . When extracting cube roots, students sometimes take all the identical prime factors instead of just one from each triple — remember only one factor comes out of each complete triple, no matter how many triples of that prime exist. Finally, do not assume every number is close to a cube; unlike squares, perfect cubes are sparse (only ten between $1$ and $1000$ ), so guessing without prime factorisation is unreliable for anything beyond the memorised table of $1$ to $10$ .

You should be able to…

State the definition of a perfect cube and recall the cubes of $1$ to $10$ without calculation.
Explain, using prime factorisation, why a number is a perfect cube exactly when every prime factor's exponent is a multiple of three.
Test any given natural number for being a perfect cube using its prime factorisation.
Find the smallest natural number by which a non-cube number must be multiplied to make it a perfect cube.
Find the smallest natural number by which a non-cube number must be divided to make it a perfect cube.
Find the cube root of a perfect cube by grouping its prime factors into triples and taking one factor from each triple.
Use the odd-number-sum pattern to express a perfect cube as a sum of consecutive odd numbers.
Solve applied problems, such as finding how many identical cuboids combine to form a perfect cube.
Recognise the Hardy-Ramanujan Number $1729$ and verify its defining property.
Distinguish clearly between the multiply-to-fix and divide-to-fix procedures for non-cubes.

Ask the tutor: Want to work through a tricky perfect-cube check or cube-root problem from your own textbook exercise together?

How is a cube root different from finding a square root?

Both are inverse operations, but they undo different things. Squaring multiplies a number by itself twice ( $n\times n = n^2$ ), so the square root asks "what number, multiplied by itself, gives this?" Cubing multiplies a number by itself three times ( $n\times n\times n = n^3$ ), so the cube root asks "what number, multiplied by itself three times, gives this?" In prime factorisation terms, a perfect square needs every prime's exponent to be a multiple of two, and you take one factor from each pair; a perfect cube needs every prime's exponent to be a multiple of three, and you take one factor from each triple. The methods are structurally identical once you replace "pairs" with "triples."

Why are there so few perfect cubes compared to perfect squares?

Cubing grows much faster than squaring. Between $1$ and $1000$ there are $31$ perfect squares (since $31^2 = 961$ and $32^2 = 1024$ ) but only $10$ perfect cubes (since $10^3 = 1000$ and $11^3 = 1331$ ). As numbers get larger, this gap widens further, because $n^3$ increases far more steeply than $n^2$ . That is also why guessing a cube root by trial is risky for large numbers — the gaps between consecutive perfect cubes become very large, so prime factorisation is the reliable method, not estimation.

If a number is not a perfect cube, is there always a smallest number to multiply it by to fix that?

Yes, always. Every natural number has a unique prime factorisation, and for each prime whose exponent is not a multiple of three, you can always supply exactly enough extra copies of that prime to push its exponent up to the next multiple of three. Multiplying by the product of all these "missing copies," one set per incomplete prime, is guaranteed to produce a perfect cube, and it is the smallest such multiplier because you added the minimum possible extra factors. The same logic works in reverse for division, except there you can only remove copies that are actually present, so division only works cleanly when the leftover count after removing full triples can be entirely divided away.

Is 1729 special only because of the taxi story, or does the pattern continue for other numbers?

The taxi story is what makes $1729$ famous, but the underlying pattern — a number expressible as a sum of two cubes in more than one way — does continue infinitely; such numbers are now called Hardy-Ramanujan numbers or taxicab numbers, and $1729$ is simply the smallest one. The next examples grow quickly, such as $4104 = 2^3+16^3 = 9^3+15^3$ and $13832 = 18^3+20^3 = 2^3+24^3$ . Finding these by hand is hard because you must search systematically through pairs of cubes and check for coincidences, which is exactly the kind of pattern-hunting Ramanujan was famous for doing in his head.