One root of a certain third-degree equation is 1. When the cubic term of the equation is crossed off, the resulting quadratic equation has a root of 2. When the squared term is also crossed off, the resulting linear equation has a root of 3. Reconstruct the original third-degree equation, expressing it in the form *ax*^{3} + *bx*^{2} + *cx* = *d*

*ax*^{3} + *bx*^{2} + *cx* + *d* = P(*x*)

We work backwards.

*cx* + *d* = 0 when *x* = 3, so 3*c* + *d* = 0, and *d* = -3*c*.

P(*x*) is now *ax*^{3} + *bx*^{2} + *cx* – 3*c*.

*bx*^{2} + *cx* – 3*c* = 0 when *x* = 2, so 4*b* + 2*c* – 3*c* = 0 → 4*b* – *c* = 0 → *c* = 4*b* and*d* = -3(4*b*) = -12*b*.

P(*x*) is now *ax*^{3} + *bx*^{2} + 4*bx* – 12*b*.

1 is a root: *a* + *b* + *c* + *d* = 0 → *a* = –*b* – *c* – *d* = –*b* – 4*b* + 12*b* = 7*b*.

Now, let *b* = 1 (or anything else we want): *a* = 7, *b* = 1, *c* = 4, *d* = -12.

So, at last, P(*x*) = 7*x*^{3} + *x*^{2} + 4*x* -12.