+483 Let \(a,b,c,d\) and \(e\) be the distinct roots of the equation \(x^5+7x^4-2=0\). Find
\( \frac{a^5}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^5}{(b - a)(b - c)(b - d)(b - e)} + \frac{c^5}{(c - a)(c - b)(c - d)(c - e)} + \frac{d^5}{(d - a)(d - b)(d - c)(d - e)} + \frac{e^5}{(e - a)(e - b)(e - c)(e - d)}. \)
Any help appreciated! 
+8 Hi, I would really want to help, but I tried to solve this problem a few times before, but I'm not quite sure if it is correct. A piece of advice is to maybe ask one of the moderators like Melody, CPhill, Alan, asinus??? Try that.
You can do this by clicking their username on the right and at the top right you can send them an email/text.
+118608 I started doing this by saying
\((x-a)(x-b)(x-c)(x-d)(x-e)=x^5+7x^4-2\)
a+b+c+d+e= -7 and abcde=2
ab +ac +ad +ae +bc +bd +be +cd +ce +de = 0
abc+abd+abe+acd+ace+ade+ bcd+bce+bde+cde = 0
abcd + abce + abde +acde + bcde =0
Maybe you can use that.
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So next I used calculus to find the turning points of \(y=x^5+7x^4-2\)
I found that there are only 2 turning points. One at x=0 and the other at x=-5.6
Since there are only 2 points there can be a maximum of 3 distinct real roots.
So the other 2 roots are not real but they are the conjugates of one another. ie p+qi and p-qi
If you let a=p+qi and b=p-qi then ab=p^2+q^2
I am just thinking in print.
Maybe you can get some ideas from this.
+483 I tried multiplying top and bottom by (x-a) and plugging in a for x in the factorization, but then you get lots of 0's in numerator and denominator, and it leads nowhere, but it seems like the right thing to do.
