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Sum of the first n cubes can be written as the square of the nth triangular numbers.

1^{3} + 2^{3} + ... + n^{3} = (1 + 2 + ... + n)^{2}.

Let us consider a number to be 63.

Now list the divisors of 63. For each list of divisors, count the number of divisors of it.

63 has 6 divisors (63, 21, 9, 7, 3, 1)

21 has 4 divisors (21, 7, 3, 1)

9 has 3 divisors (9, 3, 1)

7 has 2 divisors (7, 1)

3 has 2 divisors (3, 1)

1 has 1 divisor (1)

So sum of the cubes can be written as

6^{3} + 4^{3} + 3^{3} + 2^{3} + 2^{3} + 1^{3} = 324 = (6+4+3+2+2+1)^{2}.

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Daily Maths Topic Today - squared triangular number.