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

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Answer 1 · 2018-08-07T19:08:26

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

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#