Telescoping sum of powers 
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
 \vdots &=& &\vdots & \\
 3^3-2^3 &= &3(3^2) &-3(3) &+1\\
 2^3-1^3 &= &3(2^2) &-3(2) &+1\\ 
 1^3-0^3 &= &3(1^2) &-3(1) &+1\\  
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$
Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$
 How do this transformation was done?
Thanks!

Notice that 
$$(n-2)^3=((n-1)-1)^3=(n-1)^3-3(n-1)^2+3(n-1)-1$$
Anyway, related to the image you posted (which I didn't notice at first sight), the result was really obtained by substitution from the first row ($n\to n-1$), as other people also observed.
