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384 Answers to Exercises and Solutions to Problems

Similarly,

(Q + e)(a + f)(e + f) = -(b + c)(b + d)(c + d).

Multiplying the corresponding sides of these two equations yields the result

when (c+ d)(e+ f) # 0.

Suppose, say, that c + d = 0. If e + f = 0, too, then a = -b and the

result is clear. If e + f # 0, then (b + e)(b + f) = (u + e)(u + f) = 0 and

both sides of the required equation vanish.

Third solution. For 1 5 k 5 6, let Sk be the sum of all products of k of

the six variables and Pk be the sum of their kth powers. Define

f(x) = (x + Q)(X + b)(x + c)(x + d)(x + e)(x + f)

= (x6 + 52z4 + s4x2 + SC) + ~952.

Then f(z)-2~52: = f(- x ) , so that 0 = f(u)-2ssu = f(b)-2sgb. Therefore,

bf(a) = W4.

If ub(u + b) # 0, this equation leads directly to the result. If a = b = 0,

the result is obvious. If, say, a = 0, b # 0, then f(b) = 2s5b = 2b2cdef and

the result holds. If a + b = 0, ub # 0, then f(u) = f(b) = 0, so that ss = 0.

We can then write

f(x) = (x2 - u”)(x” - v”)(x” - w2),

where u2, v2, w2 are the zeros of t3 + szt2 + s+! + ss. Hence a = u, b = -u
fv=vw;d c, 4 e, f are, in some order, v, -v, w, -w. The result again

37. First solution. Let x = cosu+ isinu, y = cosv + isinv, z = cosw +
    i sin w, and define the symmetric functions p = x + y + z, q = xy + yz + zz,
    r = xyz. Then
    u+bi=p
    a - bi = q/r
    c+di=z2+y2+z2=p2-2q
    c - di = zs2 + ym2 + .ze2 = (q2 - 2pr)/r2.
    Since q = rjj and 121 = IyI = 1.~1 = Irl = 1, we deduce that

lq12 = IpI2 = u2 + b2.

Since 2q = (u + bi)2 - (c + di), it follows that

(u2 - b2 - c)~ + (2ub - d)2 = 4(u2 + b2).

Second solution. Let 0 = u + v + w. Then

U2 - b2 = C + z[COS(e - U) + COS(e - V) + COS(e - W)

= c+2ocosB+2bsine

2ub = d+2usinB-2bcosfI