Roots of x^3-16x^2-57x+1=0 Is a, b, and c Then a^(1/5)+b^(1/5)+c^(1/5)=?

1 Answer
Aug 7, 2018

a^(1/5)+b^(1/5)+c^(1/5) = 1

Explanation:

Here's a solution which I came to via an untidy route...

Given:

x^3-16x^2-57x+1 = 0

First use a numerical method to find the three roots of the given equation are approximately:

-3.01506549023785

0.0174583963437910

18.9976070938941

The fifth roots of these roots are approximately:

alpha = -1.246979603717467

beta = 0.4450418679126286

gamma = 1.8019377358048390

The sum of these approximate roots is very close to 1, but let's do some algebra to verify and prove.

Using the values of alpha, beta, gamma from above, find:

(x-alpha)(x-beta)(x-gamma) = x^3-x^2-2x+1

So if alpha, beta, gamma are the fifth roots of the original cubic, then x^3-x^2-2x+1 must be a factor of:

(x^5)^3-16(x^5)^2-57(x^5)+1 = x^15-16x^10-57x^5+1

We can long divide these polynomials by long dividing their coefficients thus:

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So notice that the long division is exact, confirming that the roots of x^3-x^2-2x+1 = 0 are the fifth roots of x^3-16x^2-57x+1 = 0