Cubes and cube roots

Multiply a number by itself once and you get a square; multiply it by itself twice more and you get a cube. A cube of side $5$ cm is built from $5\times5\times5 = 125$ unit cubes — that single picture is the whole idea behind this chapter. Some numbers, like $125$ , are perfect cubes: they are exactly $a^3$ for some whole number $a$ . Most numbers are not, and a large chunk of exam questions ask you to tell the two apart quickly, using nothing more than a units-digit trick and prime factorisation. This article builds up cubes and cube roots from the ground: the sign rule for negative numbers, the units-digit pattern that lets you shortlist an answer in seconds, and the triple-grouping method that turns a big number's prime factorisation into its exact cube root. Two fully worked examples — one straightforward, one with a twist — show the method end to end, and the traps section flags the two mistakes examiners see most often: stopping factorisation too early, and mixing up cubes with squares.

The core insight

A number is a perfect cube exactly when its prime factorisation splits into complete triples of identical primes, and its cube root is the product of one factor taken from each triple.

The core idea

The cube of $a$ is written $a^3 = a\times a\times a$ . The first ten cubes — $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000$ — are worth memorising cold, because recognising them on sight is faster than factorising every time.

Sign rule. Three negative signs multiplied together stay negative, so cubing a negative number always gives a negative result:

(-a)^3 = (-a)\times(-a)\times(-a) = -a^3

where: $a$ is any positive whole number, so $(-2)^3=-8$ and $(-5)^3=-125$ . This is unlike squaring, where $(-a)^2=a^2$ is always positive — cubes keep the sign of the original number, squares erase it.

Units-digit pattern. The last digit of $a^3$ depends only on the last digit of $a$ :

Units digit of $a$	0	1	2	3	4	5	6	7	8	9
Units digit of $a^3$	0	1	8	7	4	5	6	3	2	9

Digits $2$ and $8$ swap, $3$ and $7$ swap, and $0,1,4,5,6,9$ reproduce themselves. Reading this table backwards lets you shortlist the units digit of a cube root before you factorise anything — if $N$ ends in $7$ , its cube root (if it has one) must end in $3$ .

Cube root by prime factorisation. The cube root of $N$ , written $\sqrt[3]{N}$ , is the number that cubes back to $N$ . Break $N$ into prime factors, then check whether every prime occurs a multiple of three times:

\sqrt[3]{N} = \sqrt[3]{p_1^{3a}\times p_2^{3b}\times\cdots} = p_1^{a}\times p_2^{b}\times\cdots

where: $p_1, p_2, \dots$ are the distinct prime factors of $N$ , and $a, b, \dots$ are the number of complete triples each contributes. If any prime's count is not a multiple of three, $N$ is not a perfect cube.

Figure

A cube built from five layers of five by five unit cubes showing that five cubed equals one hundred twenty five

Derivation AGrouping prime factors into triples to extract a cube root

Starting from the prime factorisation of a perfect cube $N$ , show step by step why collecting the primes into groups of three and taking one representative from each group produces a number that, when cubed, exactly reconstructs $N$ — this is the reasoning that justifies the triple-grouping shortcut used on every cube-root question in this chapter.

Step 1. Write $N$ as a product of prime factors, repeating each prime as many times as it divides $N$ : $N = p_1\times p_1\times\cdots$ (some number of $p_1$ 's) $\times\, p_2\times p_2\times\cdots$ , and so on for every distinct prime dividing $N$ . This is always possible and unique for a whole number greater than $1$ , by the fundamental fact that every whole number has one prime factorisation.

Step 2. Suppose $N$ is a perfect cube, so $N = a^3$ for some whole number $a$ . Factorise $a$ itself into primes, $a = p_1^{x}\times p_2^{y}\times\cdots$ . Then $N = a^3 = p_1^{3x}\times p_2^{3y}\times\cdots$ — every prime's exponent in $N$ 's factorisation is exactly three times its exponent in $a$ 's factorisation, because cubing multiplies every exponent by $3$ .

Step 3. This means each prime's exponent in $N$ must be a multiple of $3$ , which is exactly the condition "the factors of that prime group into complete triples with none left over." Conversely, if every prime's exponent in $N$ happens to be a multiple of $3$ , set $a = p_1^{x}\times p_2^{y}\times\cdots$ where $3x, 3y,\dots$ are those exponents; then $a^3$ reconstructs $N$ exactly, so $N$ is a perfect cube.

Step 4. Taking one factor from each triple therefore builds precisely the number $a = p_1^{x}\times p_2^{y}\times\cdots$ whose cube is $N$ . This is $\sqrt[3]{N}$ , and it is the only positive whole number with that property, since cubing is strictly increasing for positive numbers (a bigger number always has a bigger cube, so no two different positive numbers can share a cube).

Worked example

Find $\sqrt[3]{1728}$ by prime factorisation, then check whether $392$ is a perfect cube.

Step 1. Factorise $1728$ down to primes: $1728 = 2\times2\times2\times2\times2\times2\times3\times3\times3 = 2^6\times3^3$ .

Step 2. Group into triples: $2^6$ splits into two triples of $2$ , and $3^3$ is already one triple of $3$ . No factor is left over, so $1728$ is a perfect cube.

Step 3. Take one factor from each triple: $\sqrt[3]{1728} = 2\times2\times3 = 12$ . Check: $12^3 = 12\times12\times12 = 144\times12 = 1728$ , which matches.

Step 4. Now factorise $392 = 2\times2\times2\times7\times7 = 2^3\times7^2$ . The $2^3$ forms a complete triple, but $7^2$ has only two $7$ 's — one short. So $392$ is not a perfect cube; multiplying by one more $7$ gives $392\times7 = 2744 = 2^3\times7^3$ , and $\sqrt[3]{2744} = 2\times7 = 14$ , since $14^3 = 2744$ .

The cube root of $1728$ is $\boxed{12}$ , and $392$ needs multiplying by $7$ to become the perfect cube $2744$ with cube root $14$ .

Competitive stretch

For a division problem instead of multiplication — "what is the smallest number to divide $392$ by to make it a perfect cube" — remove the unpaired prime instead of adding one: dividing $392 = 2^3\times7^2$ by $7^2=49$ leaves $8=2^3$ , a perfect cube with root $2$ . Deciding multiply-versus-divide always comes down to whether a prime's leftover count is closer to needing one more or to being removed entirely.

Common traps

Confusing cubes with squares. $4^3 = 64$ , not $4^2=16$ or $4\times3=12$ ; a cube always means three equal factors multiplied together, never two.
Forgetting the sign rule. Writing $(-3)^3 = 27$ by copying the square rule is wrong — an odd number of negative factors stays negative, so $(-3)^3 = -27$ .
Stopping factorisation too early. Leaving a composite factor like $4$ or $9$ ungrouped instead of breaking it down to primes ( $2, 3, 5, 7, 11,\dots$ ) makes the triple-grouping unreliable.
Grouping different primes together. A valid triple is three copies of the same prime, such as $2\times2\times2$ ; grouping $2\times3\times5$ as if it were a triple gives a wrong cube root.

Ask the tutor: Want to work through a few more perfect-cube checks together, or try the units-digit shortcut on a bigger number?

Why does a negative number cubed stay negative when a negative number squared becomes positive?

Squaring multiplies two negative factors together, and two negatives cancel to give a positive: $(-3)\times(-3)=9$ . Cubing multiplies three negative factors, and the third negative sign has nothing left to cancel it, so the result stays negative: $(-3)\times(-3)\times(-3) = 9\times(-3) = -27$ . In general, an even number of negative factors gives a positive product, and an odd number gives a negative one — cubing always involves an odd count, three.

If a number is not a perfect cube, is there still a way to find its cube root?

Not by the prime-factorisation method taught in this chapter — that method only works when every prime factor groups into complete triples, which by definition only happens for perfect cubes. For a number that is not a perfect cube, its cube root is not a whole number at all, and finding a decimal approximation needs tools beyond this syllabus, such as a calculator or logarithms. This chapter only asks you to test whether a number is a perfect cube and, if it is, to find that exact whole-number root.

Is there a quicker way than full factorisation to estimate a cube root, especially in a timed test?

Yes, for perfect cubes up to about six digits. Split the number into the last three digits and the rest; the units-digit table tells you the last digit of the root from the last digit of the number, and comparing the remaining digits to the cubes of $1$ to $9$ (memorised) tells you the tens digit. For example, for $110592$ : the last three digits $592$ end in $2$ , so the root ends in $8$ ; the remaining part $110$ lies between $4^3=64$ and $5^3=125$ , so the tens digit is $4$ , giving $48$ . This shortcut only works cleanly when the number is already known to be a perfect cube.

Cubes and cube roots

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads