SOLUTION: Find the ordered pair (a,b) of real numbers for which {{{x^2+ax+b}}} has a non-real root whose cube is 343.

First we want a cubic equation whose three roots are the three cube roots of 343.
We know that since 7³=343, that the REAL cube root of 343 is 7.  So the cubic
equation we are looking for to begin with is:
   
                             x³ = 343

whose roots are the three cube roots of 343, which are the real root 7, and the
two non-real cube roots of 343.  Get 0 on the right

                       x³ - 343 = 0

Factor:   (x - 7)(x² + 7x + 49) = 0

Use zero-factor property:

       x - 7 = 0;  x² + 7x + 49 = 0
           x = 7;  

So x² + 7x + 49 = 0 has a non-real root whose cube is 343. In fact it has
two of them!

Answer: 

 

so (a,b) = (7,49)

Edwin