Solutions to Problems; Chapter (^3 293)
(c) Cubing the three equations and adding (taking account of (a) and
(b)) yields
a3 + b3 + c3 = b3y3(z-3 + x-“) + c3z3(xm3 + yS3)
- a3x3(ys3 + z-“) + 9abc
= b3y3(-x-‘y-‘z-1 - y-3) + ,3,3(-x-~y-‘z-~ _ %-3) - a3x3(-x-‘yS1z-’ - y-“) + 9abc
= -(a” + b3 + c3) - x-‘y-‘zS1(a3x3 + b3y3 + c”z”) + Sabc,
whence the result.
8.15. Multiplying out the left side yields
3+x%(% - x) + xy(x - y) + zy(y - %) + XY(X - Y) +xz(z - X) + dy - %)
Y4Y - %> X%(% - x) “Y(X - Y)
= 3+ [(x/!/%)(x-y- t.)+(y/xz)(-x+ y- %>+(%/xY>(-x - y+ %>I
= 3+2(x3+ y3+Z3)/(xy%)= 9
by Exercise 1.5.9(a).
8.16. Let each quantity be equal to u. Then
x3 - myr = ux2 and y3 - mxz = uy2.
Subtract and divide by x - y to get
x2+xy+y2+mz=u(x+y).
Similarly, ,r2 + zy + y2 + mx = u(y + z). Subtract and divide by x - z to
getx+y+z-mmu.
8.17. Since 0 = (r + 1)3 + c(r + 1)2 + d(r + 1) + 1 = r3 + (3 + c)r2 +
(3+2c+d)r+(2+c+d), r is a zero of x3+(3+c)x2+(3+2c+d)x+(2+c+d).
Since P(x) is irreducible, this cubic must coincide with P(x) and so a =
3+c,b=3+2c+d,-1=2+c+d,whenceb=canda=3+b.Letsbe
a second zero of P(x); the third zero is l/rs, and we obtain
-3=b-a=rs+r-‘+s-‘+r+s+r-‘s-l.
Hence
0 = (r2 + r)s2 + (r2 + 3r + 1)s + (r + 1)
= [(r+l)s+l][rs+(r+l)],
from which it follows that s = -(r + 1)-l or -(r + l)r-‘.
8.18. 9 satisfies the equation
o=i+~~=(i+e)(i-e+e~-83+...-em-~+em-~).