Prime Factors in Threes Reveal a Perfect Cube

Break it into prime factors and see whether they group into threes. If every prime appears a number of times divisible by 3, it is a perfect cube; if any are left over, it is not.

That single test both identifies a cube and hands you its cube root. This page covers everything in the CBSE Class 8 Mathematics chapter's second part: cubes and perfect cubes, the factorisation test, cube roots, and making a number into a square or a cube.

The cube of a number is that number multiplied by itself three times, written $n^3$. A perfect cube is a number that is the cube of an integer.

$$3^3=3\times 3\times 3=27,5^3=125,10^3=1000$$

The perfect cubes in order are

$$1,8,27,64,125,216,343,512,729,1000$$

and these ten are worth knowing by heart, since almost every question uses one of them.

Identifying them from a list: among $27,30,64,100,125$, the perfect cubes are $27=3^3$, $64=4^3$ and $125=5^3$. The others are not.

Cubes come from volume, since a cube of side 4 cm has volume $4^3=64{\text{cm}}^3$ — which is where the name comes from.

For negative integers, the cube keeps the sign:

$$(-3{)}^3=(-3)\times (-3)\times (-3)=-27$$

That is the difference from squares worth noting. An odd index preserves a negative sign, so $(-3{)}^3=-27$ while $(-3{)}^2=+9$ — which means negative perfect cubes exist ($-8$, $-27$, $-64$) whereas a negative number is never a perfect square. Also note $64$ is both a perfect square and a perfect cube, since $64=8^2=4^3$.

Factorise into primes, then check whether each prime's count is a multiple of 3.

Worked example. Is $216$ a perfect cube?

$$216=2\times 2\times 2\times 3\times 3\times 3=2^3\times 3^3$$

Both the 2 and the 3 appear three times, so the factors split into three identical groups $(2\times 3)$. It is a perfect cube.

Worked example. Is $500$ a perfect cube?

$$500=2\times 2\times 5\times 5\times 5=2^2\times 5^3$$

The 5 appears three times, but the 2 appears only twice — not a multiple of 3. So $500$ is not a perfect cube.

Worked example. Is $3375$ a perfect cube?

$$3375=3^3\times 5^3$$

Both counts are 3, so yes.

Worked example. Is $1323$ a perfect cube?

$$1323=3^3\times 7^2$$

The 7 appears twice, so no.

The contrast with squares is exactly the useful part. A square needs its prime counts to be even — divisible by 2; a cube needs them divisible by 3. Same method, different divisor — which is why $64=2^6$ satisfies both, since 6 is divisible by 2 and by 3.

Factorise, split the primes into three identical groups, and one group is the cube root. The symbol is $\sqrt[3]{}$.

Worked example. Find $\sqrt[3]{216}$.

$$216=2^3\times 3^3$$

Take one factor from each group of three:

$$\sqrt[3]{216}=2\times 3=6$$

Check: $6^3=216$. Correct.

Worked example. Find $\sqrt[3]{3375}$.

$$3375=3^3\times 5^3\Rightarrow \sqrt[3]{3375}=3\times 5=15$$

Check: $15^3=3375$. Correct.

Worked example. Find $\sqrt[3]{2744}$.

$$2744=2^3\times 7^3\Rightarrow \sqrt[3]{2744}=2\times 7=14$$

Check: $14^3=2744$. Correct.

Worked example with a negative. Find $\sqrt[3]{-64}$. Since $64=4^3$ and the cube root of a negative is negative:

$$\sqrt[3]{-64}=-4$$

The shortcut in index form is to divide each index by 3: from $2^3\times 7^3$ the root is $2^1\times 7^1=14$. For a square root you halved the indices; for a cube root you divide them by three — and a cube root of a negative number exists, which is never true of a square root at this level.

Factorise, find which primes are short of a full group, and multiply by exactly what is missing — or divide by exactly what is spare.

To make a perfect square. Take $45$:

$$45=3^2\times 5$$

The 3 is paired but the 5 is alone. So:

- Multiply by $5$ to give $225=15^2$
- Or divide by $5$ to give $9=3^2$

To make a perfect cube. Take $392$:

$$392=2^3\times 7^2$$

The 2 forms a complete group of three, but the 7 appears only twice and needs one more. So:

- Multiply by $7$ to give $2744=14^3$
- Or divide by $7^2=49$ to give $8=2^3$

Another cube. Take $243$:

$$243=3^5$$

Five 3s is one short of six, so:

- Multiply by $3$ to give $729=9^3$
- Or divide by $3^2=9$ to give $27=3^3$

A square. Take $200=2^3\times 5^2$. The 5s are paired but there are three 2s, so multiply by $2$ to give $400=20^2$, or divide by $2$ to give $100=10^2$.

The procedure is always the same: count each prime's index, and decide how far it is from the next multiple of 2 for a square or multiple of 3 for a cube. Multiplying fills the gap and dividing removes the excess — so the two answers are different numbers, and a question asking for one will not accept the other.