Hi, I've tried the question below and got stuck halfway =) I'd really appreciate any help!! Thanks!!

QUESTION: ax^3+bx^2+cx+d=0 has roots A, B and C. Find the new equation for roots (A-k), (B-k), (C-k).

From eq given,

SOR=A+B+C=-b/a

Sum of product of roots taking 2 at a time=AB+BC+AC=c/a

POR=ABC=-d/a

Using roots (A-k), (B-k), (C-k),

SOR=A+B+C-3k=-b/a -3k

Sum of product of roots taking 2 at a time

=(A-k)(B-k)+(B-k)(C-k)+(A-k)(C-k)

=AB+BC+AC+(-A-B-B-C-A-C)k+3k^2

=AB+BC+AC+(-2A-2B-2C)k+3k^2

=c/a -2k(A+B+C)+3k^2

=c/a -2k(-b/a)+3k^2

=c/a +(2bk)/a + 3k^2

POR

=(A-k)(B-k)(C-k)

=ABC+(-AB-AC-BC)k+(A+B+C)k^2-k^3

=-d/a -k(c/a) + (b/a)k^2 - k^3

After this, I got stuck. How do I get the cubic equation? Something just seems wrong. Help! =) Thanks everyone!