>>10166278When he says "consider the LHS as a polynomial with one variable x", what he means is this: Suppose a, b, c are any three distinct numbers (like 0, 1, 2), then if you substitute those values of a, b, c in the LHS you get a polynomial in x. Then he proceeds to show that the polynomial you got by substituting the values of a, b, c is 0. That's fine. But can you conclude from this that if you algebraically expand the terms in the LHS - just using algebraic manipulation, without substituting any numbers for a, b, c - you will get 0?
For example, consider the polynomial f(x) = x(x^2 - 1). If you're working with the integers modulo 6 then f(n) = 0 for all elements of this set. But if you expand f(x), you get x^3 - x instead of 0. The reason this happens is because the set (ring, actually) of integers modulo 6 is not a field (certain multiplicative inverses do not exist).
The theory required to justify it is not really complicated but it might require some background in abstract algebra. Instead of substituting values for a, b, c and and using the resulting polynomial, you could instead form the field of rational polynomials in the indeterminates a, b, c and consider the LHS of that equation as a polynomial in x over that field. Now you can use the fact that a, b, c are 3 distinct roots and that the polynomial is of degree 2 to conclude that it must be 0.
