Proving independence among vectors: u+v If vectors u,v and w are linearly independent, will u+v, v+w, and u+w also be linearly independent?

Okay, I found a proof of this problem but, there is one key ingredient that does not quite make sense (to me).
Proof: 
(u+v)*a + (v+w)*b+(u+w)*c = 0 [because they are independant] where a,b,c are scalars
We need to prove that a=b=c=0
[This part below, I do not understand]
Since the vectors are independant:

a+b = 0
a+b =0 
b + c = 0

And solving this system we find that a=b=c=0

QED

Any help with that last part would be great, I'm not quite sure how those 3 equations were derived.
Thank you
 A: $$a\cdot(u+v)+b\cdot( v+w)+c\cdot(u+w)=(a+c)\cdot u+(a+b)\cdot v+(b+c)\cdot w$$
A: Instead of just blindly calculating, this is an excellent time to use your geometric intuition.
Consider the subspace spanned by $\{u,v,w\}$. Because they are linearly independent, $\langle u,v,w\rangle$ is a basis, and we can choose to write our vectors in that basis. Then the question reduces to whether $(1,1,0)$, $(0,1,1)$, and $(1,0,1)$ are linearly independent in $\mathbb R^3$.
Geometrically it is clear that they are -- these vectors are the diagonals of the three sides of the unit cube that meet the origin. In particular they don't all lie in a shared plane through the origin, so they span $\mathbb R^3$. And since there are only three of them, they must be linearly independent.
If you (or your instructor) don't consider "it is geometrically clear" to be a sufficiently rigorous argument, a reasonable compromise between intuition and rigor might be to compute $\begin{vmatrix}1&0&1 \\ 1 & 1& 0 \\ 0 & 1 & 1\end{vmatrix}$ and observe that it is nonzero (in fact it is 3).
