Question

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

Answer 1

`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:

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`