Math Problems
#Problems #Quest83
#Solutions | #Quest83
Any root of the equation of x³-x+1=0 satisfies the below equation :
p⁸=p⁹/p=(p-1)³/p=p²-3p+3-1/p
Thus
S=a⁸+b⁸+c⁸
S=(a²+b²+c²)-3(a+b+c)-(1/a+1/b+1/c)+9
From the equation and vieta,we can know that
a+b+c=0 , ab+bc+ac=-1 , abc=-1
Therefore
S=2-1+9=10
Author:@A_T_Uzbekistan
Any root of the equation of x³-x+1=0 satisfies the below equation :
p⁸=p⁹/p=(p-1)³/p=p²-3p+3-1/p
Thus
S=a⁸+b⁸+c⁸
S=(a²+b²+c²)-3(a+b+c)-(1/a+1/b+1/c)+9
From the equation and vieta,we can know that
a+b+c=0 , ab+bc+ac=-1 , abc=-1
Therefore
S=2-1+9=10
Author:
Math Problems
#Problems #Quest90
#Solutions | #Quest90
- Let f(x)=|x- π/6|+|x+π/3|, g(x)=arcsin(x^3-x+2)/2.
- g(x) defined, when -2 ≤ x^3 -x + 2 ≤ 2. If x > π/6, then f(x) > π/2 ≥ g(x) - no solution.
- If - π/3 ≤ x ≤ π/6, then f(x) = π2, so x^3 -x = 0, so x = 0 or x = -1.
- Only x= 0, x= -1 are solutions.
Author: @symplecticMan
- Let f(x)=|x- π/6|+|x+π/3|, g(x)=arcsin(x^3-x+2)/2.
- g(x) defined, when -2 ≤ x^3 -x + 2 ≤ 2. If x > π/6, then f(x) > π/2 ≥ g(x) - no solution.
- If - π/3 ≤ x ≤ π/6, then f(x) = π2, so x^3 -x = 0, so x = 0 or x = -1.
- Only x= 0, x= -1 are solutions.
Author:
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