Cube roots

Ask a calculator for $\sqrt[3]{13824}$ and it silently returns $24$ . Ask why $24$ , and you need a method — the same kind of systematic unwinding that turned squares into square roots, now applied one power higher. The cube root of a number answers "what number, multiplied by itself three times, gives this?", and Class 8 finds it two different ways depending on how cooperative the number is. For a perfect cube like $13824$ , prime factorisation gives an exact answer: break the number into primes, gather them into groups of three, and pull one factor from each group. For an ordinary number like $500$ , which is nobody's perfect cube, exact factorisation fails — but bracketing it between two consecutive perfect cubes still pins down its cube root to a useful estimate. This section also covers the companion skill of nudging an imperfect number into a perfect cube by multiplying or dividing away its "leftover" prime factors.

The core insight

The cube root of $n$ is the number $m$ with $m \times m \times m = n$ , found exactly for a perfect cube by grouping its prime factors into triples, and estimated for any other number by bracketing it between two consecutive perfect cubes.

The core idea

The cube root of $n$ , written $\sqrt[3]{n}$ , undoes cubing:

\sqrt[3]{n} = m \iff m \times m \times m = n

where: $n$ is the number whose cube root is wanted and $m$ is the answer — the unique number that, cubed, returns $n$ .

Perfect cubes, by grouping into triples. Write $n$ as a product of primes. If every prime's exponent is a multiple of $3$ , $n$ is a perfect cube: group the primes into triples and take one factor from each triple.

\sqrt[3]{2^3 \times 5^3} = 2 \times 5 = 10

This is the cube-root cousin of the square-root rule "pair up the primes" — but here the grouping is three at a time, not two, and mixing the two rules up is the single most common slip in this topic.

Non-perfect cubes, by bracketing. Most numbers are not perfect cubes. If

k^3 < n < (k+1)^3

then $\sqrt[3]{n}$ lies strictly between the consecutive whole numbers $k$ and $k+1$ . Comparing how close $n$ sits to $k^3$ versus $(k+1)^3$ gives a rough decimal estimate without any long-division algorithm — that machinery is outside this chapter.

Nudging a number to a perfect cube. Prime-factorise $n$ . Any prime whose exponent is not a multiple of $3$ is the obstacle: either multiply by just enough extra copies of that prime to round its exponent up to the next multiple of $3$ , or, if the exponent is small, divide out that prime's full power to remove it entirely. Both routes leave a genuine perfect cube; which one needs the smaller number is a case-by-case comparison.

Figure

Number line from 0 to 1000 marking the perfect cubes 343 and 512 with 500 sitting between them and closer to 512

Derivation ACube root of a perfect cube by prime factorisation

Show that a number whose prime factorisation has every exponent a multiple of three can be reduced to its exact cube root by grouping the prime factors into triples and keeping one factor from each triple, and confirm the result by cubing it back to recover the original number.

Step 1. Write $n$ as a product of its prime factors using repeated division, for example $n = p_1^{a_1} \times p_2^{a_2} \times \cdots$ , where each $p_i$ is prime and each $a_i$ is the number of times $p_i$ divides $n$ .

Step 2. Check every exponent $a_i$ against $3$ . If each $a_i$ is a multiple of $3$ — that is, $a_i = 3, 6, 9, \dots$ — then $n$ is a perfect cube, because $n$ can be rewritten as $(p_1^{a_1/3} \times p_2^{a_2/3} \times \cdots)^3$ .

Step 3. Read off the cube root directly from that rewritten form: $\sqrt[3]{n} = p_1^{a_1/3} \times p_2^{a_2/3} \times \cdots$ . In practice this is done by physically grouping the list of prime factors into clumps of three identical primes and taking one representative from each clump, since $p \times p \times p = p^3$ contributes exactly one factor of $p$ to the cube root.

Step 4. Verify by cubing the answer $m = \sqrt[3]{n}$ back up: compute $m \times m \times m$ and check it equals the original $n$ . This closes the loop and catches any arithmetic slip made while dividing out the primes in Step 1.

Worked example

Find $\sqrt[3]{13824}$ using prime factorisation, then estimate $\sqrt[3]{500}$ and find the smallest number that must divide $500$ to make it a perfect cube.

Step 1. Divide $13824$ by primes repeatedly: $13824 \div 2 = 6912$ , then $6912 \div 2 = 3456$ , $3456 \div 2 = 1728$ , $1728 \div 2 = 864$ , $864 \div 2 = 432$ , $432 \div 2 = 216$ , $216 \div 2 = 108$ , $108 \div 2 = 54$ , $54 \div 2 = 27$ , and then $27 \div 3 = 9$ , $9 \div 3 = 3$ , $3 \div 3 = 1$ . That is nine divisions by $2$ and three by $3$ , so $13824 = 2^9 \times 3^3$ .

Step 2. Both exponents, $9$ and $3$ , are multiples of $3$ , so $13824$ is a perfect cube. Group $2^9$ into three triples of $2$ and $3^3$ into one triple of $3$ ; taking one factor from each triple gives $\sqrt[3]{13824} = 2^3 \times 3 = 8 \times 3 = \mathbf{24}$ . Check: $24 \times 24 \times 24 = 576 \times 24 = 13824$ . ✓

Step 3. Now estimate $\sqrt[3]{500}$ . The nearby perfect cubes are $7^3 = 343$ and $8^3 = 512$ , and since $343 < 500 < 512$ , the cube root is bracketed as $7 < \sqrt[3]{500} < 8$ . Because $500$ is much closer to $512$ (distance $12$ ) than to $343$ (distance $157$ ), the estimate is a little under $8$ , roughly $7.9$ (a calculator gives $\sqrt[3]{500} \approx 7.94$ ).

Step 4. Finally, find the smallest divisor that turns $500$ into a perfect cube. Prime factorising, $500 = 2^2 \times 5^3$ . The exponent of $5$ is already $3$ , a multiple of $3$ , so $5$ needs no change. The exponent of $2$ is $2$ , not a multiple of $3$ , and since it is smaller than $3$ it cannot be topped up to the next multiple of $3$ by removing anything smaller than itself — the only clean fix is to divide out $2^2 = 4$ entirely. That gives $500 \div 4 = 125 = 5^3$ , a perfect cube, so the smallest divisor is $4$ , and $\sqrt[3]{125} = 5$ .

Competitive stretch

A quicker route to the last digit: perfect cubes have a predictable unit-digit pattern (cubes ending in $1,8,7,4,5,6,3,2,9,0$ for roots ending in $1,2,\dots,9,0$ respectively). Since $13824$ ends in $4$ , its cube root must end in $4$ — instantly narrowing the answer to a number like $24$ , $34$ , or similar, well before the full factorisation is finished.

Common traps

Grouping primes in pairs instead of triples. A number like $200 = 2^3 \times 5^2$ has a perfect-cube chunk ( $2^3$ ) but the leftover $5^2$ is not a triple, so $200$ is not a perfect cube overall — confusing the "pair up" rule for squares with the "group in threes" rule for cubes is the most frequent error here.
Bracketing with non-consecutive cubes. Always find the perfect cubes immediately above and below $n$ ; jumping to a cube two steps away gives a wider, far less useful range for the estimate.
Assuming multiplying is always the fix. When the leftover exponent on a prime is small (as with $2^2$ in $500$ ), dividing that prime out entirely is often the smaller and simpler step than multiplying up to the next triple.
Skipping the final check. Cube-root problems involve several factorisation steps in a row, so always cube the final answer back to confirm it reproduces the original number before moving on.

Ask the tutor: Want to try grouping the prime factors of a new number into triples, or practise bracketing another non-perfect cube together?

Why does grouping primes into triples work for cube roots but pairs work for square roots?

A square is a number multiplied by itself twice ( $m \times m$ ), so its prime factorisation naturally splits into pairs of identical primes — one prime from each pair rebuilds $m$ . A cube is multiplied by itself three times ( $m \times m \times m$ ), so the same logic needs groups of three identical primes, not two. The grouping size always matches the power you are undoing: square root uses pairs, cube root uses triples, a fourth root would use groups of four, and so on.

If a number is not a perfect cube, can I still find its exact cube root by prime factorisation?

No — prime factorisation only gives an exact answer when every prime's exponent is a multiple of $3$ . If even one exponent is not a multiple of $3$ , the number cannot be split into clean triples, so there is no whole-number cube root to read off. In that case Class 8 uses bracketing between two consecutive perfect cubes to estimate the value instead; getting the exact decimal cube root needs tools beyond this chapter.

How do I decide whether to multiply or divide to make a number a perfect cube?

Look at each prime's exponent that is not already a multiple of $3$ . If the exponent is small (for example $1$ or $2$ , and the prime cannot be reduced further), dividing out that prime's full power removes the obstacle entirely and is often the smaller step, as with $500 \div 4 = 125$ . If instead a small multiplier would round the exponent up to the next multiple of $3$ , multiplying is smaller. Always compare both options' sizes before picking.

Cube roots

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads