Polynomial Problem

(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8

Let's write this in a different manner

(t-1)^3  + 8  + 6(t-1)^2 + 12(t-1) 

The first two terms can be written as a sum of cubes thusly

[ (t - 1) + 2 ]  [ (t - 1)^2  - 2(t - 1)  + 4 ]  =

[ t + 1 ] [ t^2 - 2t + 1 - 2t + 2 + 4 ]  =

[ t + 1]  [  t^2 - 4t + 7 ]      (1)

The second pair of terms can be factored as

6 [ t - 1 ] [ t - 1 + 2]  =

6 [ t - 1] [t + 1]  =  [ t + 1] [6t - 6]    (2)

Putting (1) and (2) together, we have

[ t + 1 ]  [  t^2 - 4t + 7 + 6t - 6 ]  =

[ t + 1 ]  [ t^2 + 2t + 1 ]  =

[  t + 1 ] [t + 1] [t + 1 ]  =

[ t + 1]^3

Expand and simplify.

(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8

Formula:
\([1+(t-1)]^3 = t^3 = (t-1)^3 + 3(t-1)^2+3(t-1) + 1 \)

\(\begin{array}{|rcll|} \hline && \mathbf{(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8 } \\ &=& (t-1)^3 + 3(t-1)^2+3(t-1) +1 \\ && \quad + 3(t-1)^2+9(t-1)+7 \\ &=& t^3 + 3(t-1)^2+9(t-1)+7 \\ &=& t^3 + 3(t^2-2t+1)+9t-9+7 \\ &=& t^3 + 3t^2-6t+3+9t-9+7 \\ &=& t^3 + 3t^2+3t+1 \quad & | \quad (1+t)^3 = t^3+3t^2+3t+1 \\ &\mathbf{=}& \mathbf{(1+t)^3} \\ \hline \end{array} \)

thanks guys! :)