Sum of Cubes
The headline identity is that the sum of the first n cubes equals the square of the sum of the first n natural numbers: 1³ + 2³ + … + n³ = [n(n+1)/2]². In other words it is the square of the nth triangular number. So summing 1³ to 10³ is just (10×11/2)² = 55² = 3025 — no cubing needed. CAT loves this because it collapses a frightening-looking sum into one square. Two corollaries are worth memorising. First, the sum of the first n cubes is always a perfect square. Second, a partial sum from (m+1)³ to n³ is the difference of two such squares, [n(n+1)/2]² − [m(m+1)/2]². The algebraic identities a³ + b³ = (a+b)(a² − ab + b²) and, when a+b+c = 0, a³ + b³ + c³ = 3abc round out the toolkit for factor-style questions, where recognising a cube structure turns a hard expression into a quick product.
✅ Solved examples
✏️ Practice — try these, take hints as needed
📝 Topic test — 8 questions
Auto-graded with full solutions; saved to your dashboard. Use the calculator and formula sheet (top-right) any time.
Formula Reference Sheet
Cubes & cube roots
| Cube of n | n³ = n × n × n |
|---|---|
| Cube root | ∛(n³) = n |
| Perfect-cube test (prime factors) | every prime exponent is a multiple of 3 |
| Difference of cubes | a³ − b³ = (a − b)(a² + ab + b²) |
| Sum of cubes | a³ + b³ = (a + b)(a² − ab + b²) |
Sum-of-cubes identities
| Sum of first n cubes | 1³ + 2³ + … + n³ = [n(n+1)/2]² |
|---|---|
| Link to triangular number | 1³ + 2³ + … + n³ = (1 + 2 + … + n)² |
| Sum of first n cubes | = [Σn]² where Σn = n(n+1)/2 |
| Cube of a binomial | (a + b)³ = a³ + 3a²b + 3ab² + b³ |
| Identity (a+b+c=0) | a³ + b³ + c³ = 3abc |