+0  
 
+1
68
3
avatar+333 

Expand and simplify.

 

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

 

Help please. Thanks

supermanaccz  Oct 25, 2017
edited by supermanaccz  Oct 25, 2017
Sort: 

3+0 Answers

 #1
avatar+78744 
+2

(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

 

 

cool cool cool

CPhill  Oct 25, 2017
 #2
avatar+18715 
+2

Expand and simplify.

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

 

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

 

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

 

laugh

heureka  Oct 26, 2017
 #3
avatar+333 
+1

thanks guys! :)

supermanaccz  Oct 28, 2017

11 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details