Two very good solutions were sent in from Madras College, St Andrews by Dorothy and Hannah.

Imagine cubes on each of the squares marked with a blue lattice, a diagonal line of cubes of increasing size.

Then take away all the layers except the

bottom ones and spread them out, dividing

them into two groups and arranging them to cover the whole square.

Start with one cube. Then make a 3 $\times$3 $\times$1 layer by breaking the top layer of a 2 $\times$2 $\times$2 cube in half and putting the halves on the

edges.

This shows that $1^3 + 2^3 = (1 + 2)^2 $.

Then make a 6 $\times$6 $\times$1 layer by separating the layers of a 3 $\times$3 $\times$3 cube and putting the top two layers along the edges.

This shows that $1^3 + 2^3 + 3^3 = (1 + 2 + 3)^2$.

The pattern should continue. The diagram shows how

1 ³ + 2 ³ + 3 ³ + ... + 6 ³ = ( 1+ 2 + 3 + ... + 6) ² .

To prove the pattern continues indefinitely, suppose we add a new cube of side length $k$ on the bottom right hand corner. Take $(k - 1)$ layers off and divide into 2 so we have $(k - 1)/2$ layers of size $k \times k \times 1$ to be placed along two edges. Notice that if $k$ is odd then $(k - 1)/2$ is a whole number and if $k$ is even then $(k - 1)/2$ is the whole number $(k - 2)/2$ plus one half so one of the layers has to be split in half.

To find the side length L of the 'new' square, multiply $(k - 1)/2$ by $k$ to give the total length of these pieces added along the edges and add $k$ for the length of the side of the original base of the cube which is being broken up. Thus if we have $k$ cubes then the length of the sides of the large square is given by the formula.

Hence \[ 1^3 + 2^3 + 3^3 + \dots + k^3 = (1 + 2 + 3 + \dots + k)^2. \]

Footnote.

The formula for the sum of the first $k$ whole numbers is given by \[ S_k = 1 + 2 + 3 + \dots + k \] or equivalently by \[ S_k = k + (k - 1) + (k - 2 ) + \dots + 1 \] and by adding this series to itself we get $S_k = \frac{k (k +1)}{2}$.