Solutions to Problems; Chapter (^6 353)
We have that
Cw;’ = 12 - 3x11:~ = 12 - 3(24) = -60,
Cw;‘wj’ = 48 - 24Cvf2 + 9(Cv;2v;2)
= 48 - 24(24) + 9(80) = 192
and
-1
WI w2 -2w;1 =^64 - 48(Cvf2) + (CV;~V;~) - 27r1;~v;~v
= 64 - 48(24) + 36(80) - 27(64) = 64.
Hence wyl are the zeros of t3 + 60t2 + 192t - 64 and so the w; are the zeros
of 64t3 - 192t2 - 60t - 1, as required.
4.17. a, b, c, d are the zeros of the quartic t4 - wt3 - zt2 - yt -2, from which
w = u+b+c+d, z = -(ub+uc+ad+bc+bd+cd), y = ubc+abd+acd+bcd,
x = -ubcd.
4.18. First, note that Cu3 = -Cu2v = 3uvw. Let the polynomial with
zeros u, v, w be t3 +pt + q. Then
x + y + z = Cuv2 - 6uvw = 9q
xy + yz + ZX = cu3v3 + cu4vw - 3cu3v2w + 12u2v2w2
= (uv + VW + WU)~ + uvw[Cu3 - 6Cu2v + Guvw] = p” + 27q2
xyz = (-q)(-4p3 - 2792) = 4p3q + 279%
Hence x, y, z are the zeros of the polynomial
t3 - 9qP + (p3 + 27q2)t - (4p3q + 2793).
Now,
Cc3 + aCx2y + bxyz = [(9q)3 - 27q(p3 + 27q2) + 3(4p3q + 27q3)]
- a[9q(p3 + 27q2) - 3(4p3q + 27q3)] + b[4p3q + 27q3]
= 27(3 + 6u + b)q3 + (-15 - 3u + 4b)p3q.
This expression vanishes when (a, b) = (-1,3).
4.19. The cubic polynomial f(t) = (t + a)(t + b)(t + c) - x(t + b)(t + c) -
y(t + a)(t + c) - z(t + a)(t + b) h as zeros u, v, w, and so f(t) = (t - U)
(t - v)(t - w). Hence
f(-u) = -x(b - u)(c - u) = (-u - u)(-u - v)(-u - w),
so that x(a - b)(a - c) = (u + u)(a + v)(a + w). The variables y and z can
similarly be isolated.