The Art and Craft of Problem Solving

5.2 ALGEBRAIC MANIPULATION REVISITED 153

and we have g(a) = g(e) = 0, yielding the factorization

b-e e-a a-b ( b-a)( b-e)Q(a)

-a -

+

-b -

+

-e -

=

abc '

where Q(x) is a different linear polynomial. We conclude that

b-e e-a a-b ((a-b )(b-e)( e-a))

--+--+--= K ,

abc abc

for some constant K. Plugging in values (for example, a = 1, b = 2, e = 3) establishes
that K = -1. •

Example 5.2.20 (Putnam 19 39) Let x^3 + b;(2 + ex + d = 0 have integral coefficients
and roots r, s, t. Find a polynomial equation with integer coefficients written in terms
of a, b, e, d whose roots are r^3 , s^3 , t^3.

Solution: An incredibly ugly way to do this would be to solve for r, s, t, in terms
of a , b, e and then construct the cubic polynomial (x - r^3 )(x - s^3 )(x - t^3 ). Instead we
define p(x) : = x^3 + b;(2 + ex + d and note that

p(�) =0

is satisfied by x = r^3 , s^3 , t^3. We must thus convert

into an equivalent polynomial equation. Cubing comes to mind, but what should we
cube? Cubing anything but a binomial is too painful. If we put the radicals on one side
and the non-radicals on the other, we have

(1)

and now cubing both sides will remove all of the radicals. We shall employ the more
useful form of 5.2.4, which states that

(x+y)^3 =x^3 +y^3 +3xy(x+y),

and cubing both sides of (1) yields

_(x+d)^3 = (b�9)^3 + (e�)^3 + 3b�e�(b� + e�)

= b^3 ;(2 +e^3 x+3bex(b�+ e�).

On the surface, this does not look like much of an improvement, since the right-hand
side still contains radicals. But (1) allows us to substitute -(x+d) for those pesky
radicals! Our equation becomes

_(x+d)^3 = b^3 .x2 + e3x -3bex(x+d),

a cubic with integral coefficients. •