If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3 (1 + 8S1).

As the question says, we have

In this way, from (1) and (2), we have \[\mathbf{9S22}\text{ }=\text{ }\mathbf{S3}\text{ }\left( \mathbf{1}\text{ }+\text{ }\mathbf{8S1} \right).\]