310 Answers to Exercises and Solutions to Problems
8x6 - 9x3 + 1 = 0, whence x3 = 1 or l/8. The solutions (x, y) are (O,O),
(9/8,9/8), (1, w, w, 1).8.2. First, note that, if u + v = u-l + v-l, then (U + v)(l - UV) = 0,
whence u = -v or uv = 1. Applying this to u = x/a, v = b/y, we have
that, either x/u + b/y = 0 or x/a = y/b. Similar options occur for (2,~)
and (y, z). Suppose x/a + b/y = 0. Then c = Z, and a/x + b/y = 0, so that
x/u = a/x. Thus, x = fa, y = Fb. Similarly, we can handle y/b + c/z = 0
and x/a + c/z = 0.
The only remaining possibility is z/a = y/b = Z/C. If the common value
is t, then t2 - t + 2 = 0.
The complete set of solutions (x, y, Z) consists of (a, b, -c), (a, -b, c),
(-a, b, c), (ta, tb, tc) where 2t = 1 f ifi.
8.3. xy + X.Z + yz = -3ub, xyz = u3 + b3, so that x, y, z are the zeros of
t3 - 3ubt - (a” + b3) = [t - (a + b)][t2 + (a + b)t + u2 - ub + b2].8.4. x + y = z + 2, (x + Y)~ = z2 + 8 e z = 1, x + y = 3.
x3+y3-1 = 86-3~~ _ (~+y)~-3xy(x+y) = 87-3~~ u zy = -10.
(2, y,z) = (5,-2,1) or (-2,5,1).
8.5. Let u = a2 - x2 = b2 - y2 = c2 - z2. Then
u - &Gk/~=&T&Lzi+~iq
j [2u2 - (a2 + b2)u + a2b2] - 2udzdG
= (c” - u)(u2 + b2 - 2u + 2d=&5-i)
j 2c2u + u2b2 - c2(a2 + b2) = 2c2Ja2b2 - (a2 + b2)u + u2
+ 4a2b2c2u + u4b4 + a4c4 + b4c4- 2u2b2c2(a2 + b2 + c2) = 0
u2 + b2 + c2
a u=
(^2 1)
(x,y,z) = f;(bcu-’ - abc-’ - acb-l) ,
+acb-’ - abc-’ - bcu-l), f;(ubc-’ - acb-’ - bca-l)).
The task of checking these solutions is left to the reader.
8.6. Multiplying the second equation by 14xy and adding the result to the
first yields
-7y4 + 14x3y - 67x2y2 + 56xys +^64 =^0
3 -7y4 + 56xy3 - 67x2y2 + 14x3y + (x2 - 7xy + 4~~)~ =^0
3 x4 - lOx2y2 + 9y4 = 0
- (x - 3y)(x + 3y)(x - y)(x + y) = 0.