Using polynomial $x^n+ax+b=0$ and proving an equation

Question

If $\ a_1$, $\ a_2$, $\ a_3$,..., $\ a_n$ are roots of the equation $x^n+ax+b=0$, then prove that $(\ a_1-\ a_2)(\ a_1-\ a_3)(\ a_1-\ a_4)..(\ a_1-\ a_n)=n\ a_1^{n-1}+a$.

By taking n=3 and using the roots 2,4 &-6 i am able to prove the answer but not able to do via substitution.

Answer 1

Since $a_1 \ldots a_n$ are roots of $x^n + ax + b$, then $$x^n + ax + b = (x-a_1)\ldots(x-a_n)$$ Derive both sides w.r.t $x$ you get: \begin{equation} \begin{split} nx^{n-1} + a &= (x-a_2)(x-a_3)\ldots(x-a_n) \\ &+ (x-a_1)(x-a_3)\ldots(x-a_n) \\ &+ (x-a_1)(x-a_2)\ldots(x-a_n) \\ &+ \vdots \\ & + (x-a_1)(x-a_2)\ldots(x-a_{n-1}) \end{split} \end{equation} Plug $x = a_1$ on both sides you get \begin{equation} \begin{split} na_1^{n-1} + a &= (a_1-a_2)(a_1-a_3)\ldots(a_1-a_n) \\ &+ (a_1-a_1)(x-a_3)\ldots(a_1-a_n) \\ &+ (a_1-a_1)(x-a_2)\ldots(a_1-a_n) \\ &+ \vdots \\ & + (a_1-a_1)(a_1-a_2)\ldots(a_1-a_{n-1}) \end{split} \end{equation} Notice that all terms from the second to the last rows above are zeros due to $a_1-a_1$, hence we get $$na_1^{n-1} + a = (a_1-a_2)(a_1-a_3)\ldots(a_1-a_n)$$

Answer 2

If you are using $a$ as a coefficient, then using $a_j$ for a typical root is not a good choice of notation.

You might want to observe that if $f(x)=x^n+ax+b$, then $f'(x)= nx^{n-1}+a$, so your RHS is $f'(a_1)$. Is that significant?