🇺🇸 USA Math Olympiad Algebra Problem | Find the Value of x

[@suchitkumar5990]

Superb sir ❤❤❤❤

[@ManojkantSamal]

X=76..(May be ) ^=read as to the power *=read as square root As per question {X+*(x^2+32)}^(1/5)+{x- *(x^2+32)}^(1/5)=2 Let a^5={x+*(x^2+32)} b^5={x-*(x^2+32)} So, a+b=2.......eqn1 a^5+b^5=x+*(x^2+32)+x-*(x^2+32) =2x........ Eqn2 a^5×b^5={x+*(x^2+32)}×{x-(*x^2+32)} =x^2 -(x^2+32) =x^2-x^2-32=-32 ab=(-32)^(1/5)= -2.......eqn3 Now (a+b)^5=a^5+b^5+5ab{2ab(a+b)+b^3+a^3} So, 2x+5ab{2ab(a+b)+(a+b)^3-3ab(a+b)} =2^5 5ab(a+b){2ab+(a+b)^2-3ab}=32-2x 5ab(a+b){(a+b)^2-ab}=32-2x (10×-2){2^2-(-2)}=32-2x (-20)(4+2)=32-2x (-20)×6=32-2x 32-2x=-120 -2x=-120-32=-152 X=-152/-2=76 Hence X=76

[@kareolaussen819]

Note that Wolfram Alpha does not return 2 from the following input (76+√(78^2+32)^(1/5) + (76-√(76^2+32))^(1/5) = 3.32... + 0.43.. i However (√(76^2 +32)+76)^(1/5) - (√(76^2+32)-76)^(1/5) = 2. The odd rooats of negative numbers or not usually defined to be negative, but rather complex quantities.