Alterative Sum of Squared Error formula proof The well-known formula of calculating Sum of Squared Error for a cluster is this:
SSE formula
where "c" is the mean and "x" is the value of an observation.
But this formula also brings the same result:
Alternative SSE formula
where "m" is the number of the observations and "y" takes in every iteration, values of the observations.
For example, if we have {3, 7, 8} , our mean "c" = 6 and:
Using the usual formula: (6-3)² + (6-7)² + (6-8)² = 14
Using the alternative formula: [ 1∕(2*3) ] × [ (3-3)² + (3-7)² + (3-8)² + (7-3)² + (7-7)² + (7-8)² + (8-3)² + (8-7)² + (8-8)²] = 14 
Starting from the first formula, I 'm trying to prove the alternative, but I 'm lost. Can someone help me with the proof?

Working with your numerical example:
$$14=SSE=(6-3)² + (6-7)² + (6-8)² = \\
\left(\frac{3+7+8}{3}-3\right)^2+\left(\frac{3+7+8}{3}-7\right)^2+\left(\frac{3+7+8}{3}-8\right)^2=\\
\frac{(7-3+8-3)^2}{3^2}+\frac{(3-7+8-7)^2}{3^2}+\frac{(3-8+7-8)^2}{3^2}=\\
\frac{2[(7-3)^2+(8-3)^2+(7-8)^2]+2[(7-3)(8-3)+(3-7)(8-7)+(3-8)(7-8)]}{3^2}=\\
\frac{3[(7-3)^2+(8-3)^2+(7-8)^2]}{3^2}-\frac{[(7-3)+(3-8)+(8-7)]^2}{3^2}=\\
\frac{2[(7-3)^2+(8-3)^2+(7-8)^2]}{2\cdot 3}-0=\\
\frac{\sum_{i}\sum_{j}dist(x_i,y_j)^2}{2\cdot3}.$$
