Hi, spin_rgwga:
Since your problem involves (small) powers of a,b,c, we will have
occasion below to refer them as "bases".
The short answer is that the three by three linear system in unknowns x,y,z:
x + ay + a^2z = d
x + by + b^2z = e
x + cy + c^2z = f
has a unique solution precisely when the numbers a,b,c are all
distinct. We will mention several ways to establish this.
An elementary approach to showing this is by considering the
equivalent matrix formulation:
/ 1 a a^2 \ / x \ / d \
 1 b b^2   y  =  e 
\ 1 c c^2 / \ z / \ f /
which by standard results of linear algebra has a unique solution just
when the matrix of coefficients, involving powers of a,b,c, is
invertible (nonsingular), or equivalently when the determinant of that
coefficient matrix is nonzero:
 1 a a^2 
 1 b b^2  = (ba)(ca)(cb)
 1 c c^2 
as you can easily establish for yourself by expanding the determinant
or using elementary row operations to put it in upper triangular form.
[See below for one easy way to carry out this computation.]
Vandermonde matrices: the general case
======================================
This 3x3 matrix is one case of a more general class of Vandermonde
matrices, whose determinants are the always product of differences
between the "bases" taken in an order corresponding to the order of
the rows in which each appears:
[The Vandermonde Matrix]
http://wwwmath.cudenver.edu/~rrosterm/crypt_proj/node6.html
It should be clear that the determinant is nonzero (and thus the
matrix is invertible and the solution of the linear system is unique)
exactly when all the bases are distinct (here a,b,c all different),
for the product is zero if and only if one of the factors is zero
(implying that two of the bases are equal).
An equivalence with polynomial interpolation
============================================
There is an alternative way to think about the problem which leads to
this same conclusion. Suppose you are asked to find coefficients of a
polynomial p of degree at most 2:
p(u) = z*u^2 + y*u + z
such that p(a) = d, p(b) = e, p(c) = f. Such a problem,
"interpolating" the values at three prescribed points u = a,b,c, will
have a unique solution whenever the points a,b,c are all distinct.
[Note that if two points out of a,b,c were the same, we could force no
solution to exist by requiring the corresponding right hand sides out
of d,e,f to differ. Likewise, if two "bases" and their right hands
sides both agree, then we really have at most two independent
conditions, not enough information to uniquely determine all three
coefficients of polynomial p.]
Computing the determinant
=========================
You may wish to confirm the determinant calculation with Maple, which
handles symbolic expressions. However the elementary row operations
of adding a multiple of one row to another do not change the
determinant. Therefore the following is easily carried out "by hand":
 1 a a^2   1 a a^2 
 1 b b^2  =  0 ba b^2a^2 [subtract top row from 2nd & 3rd rows]
 1 c c^2   0 ca c^2a^2
Now remove a factor of (ba) from the 2nd row and (ca) from the 3rd:
 1 a a^2   1 a a^2 
 1 b b^2  = (ba)(ca)  0 1 b+a 
 1 c c^2   0 1 c+a 
 1 a a^2 
= (ba)(ca)  0 1 b+a  [sub. 2nd from 3rd row]
 0 0 cb 
= (ba)(ca)(cb)
regards, mathtalkga 
Clarification of Answer by
mathtalkga
on
05 Feb 2005 19:08 PST
Some additional remarks about actually solving the linear system,
versus showing under what conditions the solution is unique.
When the determinant is nonzero, one can invert the coefficient matrix:
/ 1 a a^2 \
 1 b b^2 
\ 1 c c^2 /
symbolically (ie. with Maple or other computer algebra program) and
get a result like this:
/ 2 2 2 2 2 2 \
 b c  b c a c  a c a b  a b 
1  
  2 2 2 2 2 2 
(ba)(ca)(cb)  b  c c  a a  b 
 
 
\ c  b a  c b  a /
Then multiplying the "constant" terms of the right hand side:
/ d \
 e 
\ f /
by the above 3x3 matrix inverse gives the unique solution for x,y,z.
As a practical matter, however, linear systems are rarely solved using
a matrix inverse (though such a small system as this might be counted
among those exceptions).
In particular the matrix inverse approach to interpolation problems
suffers from numerical disadvantages when "nodes" (bases) a,b,c are
particularly close to one another (so that dividing by their product
leads to large entries in the matrix inverse).
Another approach is to take a sum of polynomials which are constructed
for interpolating a fixed set of nodes (or bases, as we've been
calling a,b,c).
Consider for example these quadratics in variable u:
(ub)*(uc)
A(u) = 
(ab)*(ac)
(ua)*(uc)
B(u) = 
(ba)*(bc)
(ua)*(ub)
C(u) = 
(ca)*(cb)
These three polynomials have the characteristics of being 1 at exactly
one of the three nodes and 0 at the other two. Thus we can explicitly
write the interpolating quadratic for our data this way:
p(u) = d*A(u) + e*B(u) + f*C(u)
The linear dependence of p(u) on the constant terms d,e,f is apparent.
The general case of these is called Lagrange interpolating polynomials. See:
[Lagrange Interpolating Polynomial  from MathWorld]
(Eric W. Weisstein  A Wolfram Web Resource)
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html
regards, mathtalkga
