tk8 prob 6

No it is not correct.    

I think i did it the hard way,....

(x-1)(x-2)(x-3)=(6-1)(6-2)(6-3)   

So x=6 is one solution.  

It is a cubic so there is a total of 3 answers for x.   

The other 2 satisfy   x^2+k=0

Maybe there is a quicker way to do it but I expanded the left.

$(x-1)(x-2)(x-3) =5*4*3\\ x^3-6x^2+11x-6=60\\ x^3-6x^2+11x-66=0\\ \text{Then I divided the LHS by (x-6) and got}\\ (x-6)(x^2+11)=0\\ So \\ x=6,\;and\;\;x^2=-11\\ x^2=-11\\ x=\pm\sqrt{-11}\\ x=x=\pm\sqrt{11}\;i\\ $

But you were not asked for all the x solutions you were asked for k and  k=11