An algebra problem by suresh jh

What you have on the LHS (other than the $x^3$ term is what's called a Lagrange Interpolation equation of $y = x^3$ given $n$ points along the curve at $x_i=\alpha_i, y_i = \alpha_i^3$ .

Since the interpolation is of polynomial order $n-1$ and it interpolates a cubic, which can be achieved with at most a cubic. The interpolation is over determined and as a result an infinite number of $x$ 's will satisfy the equation.

All $a_i$ must be distinct or else there will be division by zero somewhere in the expression. Explanation: From the denominator of the leftmost additive term, $\prod_{i=2}^n (\alpha_1-\alpha_i)$ , we can immediately see that $\alpha_1$ cannot be equal to any other $\alpha_k$ (i.e., $2\leq k\leq n$ ) because that would lead to division by zero. In particular, $\alpha_1\neq \alpha_n$ . Now, we can direct our attention to the denominators produced by the summation, which are of the form: $\prod_{i\neq j} (\alpha_j-\alpha_i)$ , where $j$ ranges from $2$ to $n-1$ . From this, we can see that if $\alpha_k=\alpha_i$ and $k\neq i$ for any $k,i$ such that $2\leq k \leq n-1$ and $1\leq i \leq n$ , then one of those denominators will be zero. Summing up these two conclusions, we deduce $\alpha_i\neq \alpha_j$ for any pair $i,j$ such that $i\neq j$ .

If one substitutes $a_i$ for all $1\leq i \leq n$ , then they are all zeros of the polynomial. Explanation: First, take $x=\alpha_1$ . The entire expression reduces to $\alpha_1^3+0+0-\alpha_1^3=0$ . Next take $x=\alpha_k$ such that $2\leq k\leq n-1$ . The entire expression reduces to $0+\alpha_k^3+0-\alpha_k^3=0$ . Finally, taking $x=\alpha_n$ , we get $0+0+\alpha_n^3-\alpha_n^3=0$ . Therefore, $\alpha_1,\alpha_2,...,\alpha_n$ are all roots of the equation.

Since a polynomial of degree $n-1$ or less has $n$ or more distinct zeros, then it must be the zero polynomial. Therefore, any $x$ is a root of the equation.