chapter  7
Polynomial Equations
Pages 10

Consider the cubic equation x3 + ax2 + bx + c = 0 (7.1)

The first step is to get rid of the x2 term. This is easily done: put y = x+ a3 . Theny3 = (x+ a3 )3 = x3 +ax2 + a23 x+ a327 , so Equation (7.1) becomes y3 +b′y+c′ = 0for some b′,c′. Write this equation as y3 + 3hy + k = 0 (7.2)

(The coefficients 3h and k can easily be worked out, given a,b,c.)Here comes the clever part. Write y = u + v. Then y3 = (u + v)3 = u3 + v3 + 3u2v + 3uv2 = u3 + v3 + 3uv(u + v)

= u3 + v3 + 3uvy . Hence, the cubic equation

y3−3uvy− (u3 + v3) = 0 (7.3) has u + v as a root.Our aim now is to find u and v so that the coefficients in Equations (7.2) and(7.3) are matched up. To match the coefficients, we require

h =−uv, k =−(u3 + v3) . (7.4) From the first of these equations we have v3 = −h3u3 , hence the second equationgives u3− h3u3 =−k, so u6 + ku3−h3 = 0 . (7.5) This is just a quadratic equation for u3, and a solution is

u3 = 12 ( −k +√k2 + 4h3) .