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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.
First, you need to read, to interpret and to understand the condition correctly.
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| The problems asks to find the maximal value of the expression of r^3 - s - t |
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| over ALL POSSIBLE PERMUTATIONS of the roots r, s and t. |
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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, and ,
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.
ANSWER. Under given conditions, the maximal value of the expression r^3 - s - t is 2.
Solved (correctly).