1. ## Sum of squares?

[(1+t)^3 +3(1+t)^2 +2(1+t)] -6
(1+t) -1

So I know the sum a cube gives me (1+t)(1-t+t^2) for the first part of the problem, right?
I am having difficulty with 3(1+t)^2 since I know there is no sum of squares; how am I supposed to factor/multiply this?

2. $\frac{(1+t)^{3}+3(1+t)^{2}+2(1+t)-6}{(1+t)-1}$

Expand out the numerator, and it is obvious the denominator is just t.

The numerator simplifies and factors to $t(t^{2}+6t+11)$

When we divide by t we get $t^{2}+6t+11$

3. Would you mind showing me how you got t(t^2+6t+11) ?
I keep getting way too many numbers that cannot be simplied or cancelled.

4. *simplified, sorry

5. The numerator simplifies to $t^{3}+6t^{2}+11t$

Then, I just factored out a t.

6. Originally Posted by galactus
The numerator simplifies to $t^{3}+6t^{2}+11t$

Then, I just factored out a t.
I'm sorry but I really just don't understand this. Can you possibly show me what you did from the beginning of the problem? I just don't see how it comes out $t^{3}+6t^{2}+11t$
Thank you, I really do appreciate your time and sorry for all the questions.

7. It's nothing fancy, just expand it out.

$(1+t)^{3}=t^{3}+3t^{2}+3t+1$

$3(1+t)^{2}=3t^{2}+6t+3$

$2(1+t)=2+2t$

Put it all together:

$t^{3}+3t^{2}+3t+1+3t^{2}+6t+3+2+2t-6$

Now, add like terms and simplify:

$t^{3}+6t^{2}+11t$

There, that's all there is to it.