Problem2_Equation

x>0 
y>0
x,y both belong to real numbers

let y^2 =t
solution to problem lies in finding a pair (t,x) such that 
t^3= x^6 + 8*(x^4) - 6*(x^2) +8



(SOLUTION IN PYTHON3:)

                              def is_perfect_cube(x):
                                  return int(round(x**(1./3))) ** 3 ==x
    
                              def calc():
                                  for i in range(100):
                                       y= (i**6) + 8*(i**4) -6*(i**2) +8
                                       if(is_perfect_cube(y)):
                                           print(i)
          
          
          

the above code checks for x in range 0 -100 if it satisfies the above equation
and the first solution is found for x=3 and t=11

so answer 

3345x + 4321t= 3345*3 + 4321*11=57566

NOTE: ITS GIVEN THAT  SOLUTION DOES NOT EXCEED 10^7 SO FEASIBLE RANGE FOR CHECKING (BRUTE – FORCING)  FOR THE VALUE OF x WAS TILL 100.