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Solutions to Problems; Chapter 4 313

8.14. From the last two equations

(y + %)2 h (y - z)2 + 4 y% = 914 - 3x + x2 + 314 = x2 - 3x + 3.

Hence u3 = x(x2 - 3x + 3) -^1 = (Z - 1)3, so that x =^1 + a,^1 + wu or
1 + w2a, where w is an imaginary cube root of unity. Substituting this into
the last two equations yields quadratic equations for y and Z.

8.15. Adding and subtracting the two equations yields

2(x2 - y2) = (u + b)x - (a - b)y

4xy = (a - b)x + (a + b)y.

Hence

and

2(x + ~i)~ = [(a + b) + (a - b)i](x + yi)

2(x - ~i)~ = [(a + b) - (a - b)i](x - yi).
If real solutions x, y are sought, these equations are equivalent. We obtain
x + yi = Cl whence (x, y) = (0,O) or 2(x + yi) = (a + b) + (u - b)i, whence

(2, Y> = ((a + b)P, (a - bY2).
However x and y may be nonreal and a more careful analysis is needed.
If x + yi and x - yi are both nonzero, then we are led to the second solution
above. The remaining cases are as follows:
(1) x + yi = 0 and 2(x - y;) = (a + b) - (a - b)i

* (x7 Y) =

(

(a + b) - (u - b)i (a - b) + (u + b)i

4 ’ (^4) > ;
(2) x - yi = 0 and 2(x + yi) = (a + b) + (u - b)i

• (2, Y) =
  (
  (u + b) + (a - b)i (u - b) - (a + b)i
  4 ’ (^4) >
  ,
  8.16. Equating two expressions for b2c2 yields
  0 = (v2+u2 +%“)( w2 + y2 + u”) - [VW + u(y + z)]”
  = (uw - vy)2 + (uv - W%)2 + (u” - yz)2.
  Hence, uw = vy, uv = wz, u2 = ye, and similarly, VW = UX, v2 = XZ,
  w2 = xy.
  By substitution, we obtain x(x + y + Z) = a2, y(x + y + z) = b2,
  %(X + y + %) = c2, whence (x + y + z)~ = a2 + b2 + c2 = d2. Therefore
  (x, y, z, u, v, w) = *(a2/d, b2/d, c2/d, bc/d,ca/d, ah/d).
  8.17. Eliminating u from each adjacent pair of equations yields
  v(y - x) = b - ux