Question 1178729
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Let r, s, and t be the roots of the equation x^3 - 2x + 1 = 0 in some order. 
What is the maximal value of r^3 - s- t?
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            The solution in the post by  @MathLover1  is not sufficient.


            I came to bring the correct solution.



<pre>
First, you need to read, to interpret and to understand the condition correctly.


    +------------------------------------------------------------------------------------+
    |    The problems asks to find the maximal value of the expression of  r^3 - s - t   |
    |                                                                                    |
    |    over ALL POSSIBLE PERMUTATIONS of the roots r, s and t.                         |
    +------------------------------------------------------------------------------------+



To find the value of  r^3 - s - t  for only one possible permutation, as @MathLover1 does,
IS NOT ENOUGHT to solve the problem.



Now, if r is one of the roots, then  r^3 - 2r + 1 = 0,  which implies


    r^3 = 2r - 1  and further  r^3 - s - t = (2r-1) - s - t = 3r - (r + s + t) - 1.


The sum of the roots  (r + s + t) is the coefficient at x^2 of the original equation, taken with
the opposite sign (the Vieta's theorem).

In our case, this coefficient is zero;  therefore


    r^3 - s - t = 3r - (r + s + t) - 1 = 3r - 1.


THEREFORE,  the expression  r^3 - s - t  is maximal when 3r  is maximal, or, equivalently,  
when the root  "r"  is maximal of the three roots.


The roots of the equation  x^3 - 2x + 1 = 0  are  1,  {{{(-1+sqrt(5))/2}}}  and  {{{-(1+sqrt(5))/2}}},

    as @MathLover1 did find in her post.


Of them, the root  1  has maximum value.

THEREFORE, from what is written above in my post, the maximal value of the expression  r^3 - s - t  is  3*1 - 1 = 2.   


<U>ANSWER</U>.  Under given conditions, the maximal value of the expression  r^3 - s - t  is  2.
</pre>

Solved (correctly).