Lagrange interpolation formula question

Question

Find the number of values of $x$ satisfying the relation $$ \alpha_1^3 \left( \frac{\prod_{i=2}^n (x - \alpha_i)}{\prod_{i=2}^n (\alpha_1 - \alpha_i)} \right) + \sum_{j=2}^{n-1} \left( \left( \frac{\prod_{i=1}^{j-1}(x-\alpha_i)\prod_{i=j+1}^n(x-\alpha_i)}{\prod_{i=1}^{j-1}(\alpha_j - \alpha_i) \prod_{i=j+1}^n(\alpha_j - \alpha_i) } \right) \alpha_j^3 \right) + \left( \frac{\prod_{i=1}^{n-1}(x-\alpha_i)}{\prod_{i=1}^{n-1}(\alpha_n - \alpha_i)} \right)\alpha_n^3 - x^3 = 0 $$ where $n \geq 5$

Answer 1

In your notation, $p(x)$ is a Lagrange interpolation of degree $n-1$, hence $q(x)$ is of degree $n-1 > 3$. Therefore, if $q(x)$ is not the zero polynomial, it can have at most $n-1$ roots for $q(x) = 0$.

On the other hand, because $p(\alpha_i) = \alpha_i^3$, you already know that $$\{\alpha_i\mid i = 1,2, \dots, n\}$$ are all roots of $q(x)$. Hence, if $\alpha_i$'s are all distinct, there are at least $n$ roots for $q(x)$.

Therefore, $q(x)$ has more roots than its degree, ergo $q(x) = 0$ and there are infinite number of roots.

Answer 2

Below I copied the relevant part of my answer to the same question.

It is easy to see that $p(x)$ is the interpolation polynomial in the Lagrange form of a function $x^3$. That is, $p(x)=x^3$ when $x=\alpha_i$ for each $i$. Since $q(x)$ is a polynomial of degree $n-1$ which has at least $n$ roots $\alpha_i$, $q(x)$ is the zero polynomial. That is, the initial equality holds for each $x\in\Bbb C$.

If you wish, I can provide for you other answers from this deleted thread.

Answer 3

$$\alpha_1^3 \left( \frac{\prod_{i=2}^n (x - \alpha_i)}{\prod_{i=2}^n (\alpha_1 - \alpha_i)} \right) + \sum_{j=2}^{n-1} \left( \left( \frac{\prod_{i=1}^{j-1}(x-\alpha_i)\prod_{i=j+1}^n(x-\alpha_i)}{\prod_{i=1}^{j-1}(\alpha_j - \alpha_i) \prod_{i=j+1}^n(\alpha_j - \alpha_i) } \right) \alpha_j^3 \right) + \left( \frac{\prod_{i=1}^{n-1}(x-\alpha_i)}{\prod_{i=1}^{n-1}(\alpha_n - \alpha_i)} \right)\alpha_n^3 =x^3$$

$$\alpha_1^3 \left( \frac{(x - \alpha_2)(x - \alpha_3)......(x - \alpha_n)}{(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3)......(\alpha_1 - \alpha_n)} \right) +\alpha_2^3 \left( \frac{(x - \alpha_1)(x - \alpha_3)......(x - \alpha_n)}{(\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3)......(\alpha_2 - \alpha_n)} \right) +\alpha_3^3 \left( \frac{(x - \alpha_1)(x - \alpha_2)......(x - \alpha_n)}{(\alpha_3 - \alpha_1)(\alpha_3 - \alpha_2)......(\alpha_2 - \alpha_n)} \right) +.........+........+\alpha_n^3 \left( \frac{(x - \alpha_1)(x - \alpha_2)....(x - \alpha_{n-1})}{(\alpha_n - \alpha_1)(\alpha_n - \alpha_2)....(\alpha_n - \alpha_{n-1})} \right)=x^3$$ Given equation has degree $n-1$ which satisfy $\alpha_1,\alpha_2,\alpha_3,........\alpha_n$ so it become a identity.
Hence, it has infinity many solution.