can't do the question right now but i assume that this is how you do it:
since x^3 + 3ax^2 + 3bx + c = 0 has a double root then this means that p(x) = x^3 + 3ax^2 + 3bx + c has a double zero. let this double zero be y. then p(y) = 0 and p'(y) = 0. this means that y^3 + 3ay^2 + 3by + c = 0 and 3y^2 + 6ay + 3b = 0 -> y^2 + 2ay + b = 0. now we solve this last equation for the double root, y. u will get 2 solutions (since quadratic), then sub each presumed double root into the original equation (p(y)). whichever results in p(y) = 0 is the double root. good luck!
Notice that the required answer includes the constant term . That should be a hint - if we were to differentiate the polynomial, we would lose , so not only would we have to expand the cube of a binomial (which can be really messy), we would also have to rewrite the result to include in order to find the double root, so let's try a different approach.
Let and let the roots be .
By the relationships between the roots and co-efficients,