Polynomial uniqueness proof this problem is from my Gelfand's Algebra book.
Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.
This means that if $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2 and $P(x_1)=Q(x_1), P(x_2)=Q(x_2), P(x_3)=Q(x_3)$ for three different numbers $x_1, x_2,$ and $x_3$, then the polynomials $P(x)$ and $Q(x)$ are equal.
I honestly don't know where to start.
 A: A polynomial of degree two is defined by three of it's values. Why? If we want to obtain a $2^{\rm nd}$ degree polynomial, we must specify three constants, $a_1,a_2,a_3$. But what are this constants? Say
$$p(x)=a_1x^2+a_2 x+a_3$$
Then, knowing three values, $b_1,b_2,b_3$ of points $x_1,x_2,x_3$, we have that
$$\begin{cases} a_1x_1^2+a_2 x_1+a_3=b_1\\ {}\\a_1x_2^2+a_2 x_2+a_3=b_2\\{}\\a_1x_3^2+a_2 x_3+a_3=b_3\end{cases}$$
and this is a system of three linear equations in three unknowns $a_1,a_2,a_3$, which we can always solve (meaning we can always determine its solutions, which might be none at all). This generalizes: an $n$ degree polynomial is uniquely determined by $n+1$ of its values. 
A: Hint $\rm\:f(x) = ax^2\!+\!bx\!+\!c,\ a\ne 0\:$ and $\rm\:f(x) = f(y) = f(z)\:$ implies the following determinant $= 0$, having proportional first and last columns
$$\rm 0\, =\, \left | \begin{array}{ccc} 
\rm f(x) & \rm x & 1 \\
\rm f(y) & \rm y & 1 \\
\rm f(z) & \rm z & 1
\end{array} \right | 
\, =\,
\left | \begin{array}{ccc} 
\rm ax^2 & \rm x & 1 \\
\rm ay^2 & \rm y & 1 \\
\rm az^2 & \rm z & 1
\end{array} \right |  \, =\, a\,(x-y)(y - z)(z-x) $$
Thus, if the coefficient ring is a field (or domain), one of the RHS factors must be $0$. Since $\rm\:a\ne 0\:$ one of the other factors $= 0,\:$ i.e. the roots are not distinct.
Remark $ $ That the second determinant equals the first follows from the fact that the first columns are congruent modulo the others. More precisely, the second det arises by applying the elementary col operation $\rm\: col_1 \leftarrow   col_1 - b\cdot col_2 - c\cdot col_3,\:$ yielding $\rm\: f(x) - b\cdot x - c\cdot 1 = ax^2,\:$ etc. The same argument works for higher degree polynomials. For the general case, it helps to know that the  rhs matrix, after removing the factor of $\rm\:a,\:$ is the ubiquitous Vandermonde matrix, 
A: Hint: $P-Q$ is a polynomial of degree not exceeding $2$ and has at least three roots $x_1, x_2, x_3$. How many roots would a nonzero polynomial of such degree have?
