Page 47 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Proof: Let F(n) = a + b + c . Obviously a, b, c are roots of the equation
n
n
n
(x − a)(x − b)(x − c) =0. This equation is equivalent to
x =(a + b + c)x − (ab + bc + ca)x + abc .
2
3
n
When n ≥ 4, we have x =(a + b + c)x n−1 − (ab + bc + ca)x n−2 +(abc)x n−3 .
n
Thus a =(a + b + c)a n−1 − (ab + bc + ca)a n−2 +(abc)a n−3 .
n
Similarly, b =(a + b + c)b n−1 − (ab + bc + ca)b n−2 +(abc)b n−3
n
and c =(a + b + c)c n−1 − (ab + bc + ca)c n−2 +(abc)c n−3 . Add the above three equalities
together: F(n) =(a + b + c)F(n − 1) − (ab + bc + ca)F(n − 2) + (abc)F(n − 3).

In addition, we know a + b + c +2ab +2bc +2ca =(a + b + c)
2
2
2
2
and a + b + c − 3abc =(a + b + c)(a + b + c − ab − bc − ca). When a + b + c =0,
3
3
3
2
2
2
2
2
2
a +b +c
1
1
1
3
3
3
(1) = 0, ab + bc + ca = − 2 = − F(2), abc = (a + b + c )= F(3),F(n)=
3
3
2
1
1
then F(2)F(n − 2) + F(3)F(n − 3)
2 3
2
2
a +b +c 2 1 1 3 3 3 1
= − F(2), abc = (a + b + c )= F(3),F(n)=
2
(1) = 0, ab + bc + ca = − a +b +c 2 2 1 3 1 3 3 3 3 1 a +b +c 2 1 1 3 3 3 1
2
2
2
2
(1) = 0, ab + bc + ca =
(1) = 0, ab + bc + ca = −
1
1 F(2)F(n − 2) + F(3)F(n − 3) 2 = − F(2), abc = (a + b + c )= F(3),F(n)= − 2 = − F(2), abc = (a + b + c )= F(3),F(n)=
3
3
2
2
3
3
1
(4) = F (2). Choose
2
1
1
1
2 1 F(2)F(n − 2) + F(3)F(n − 3) . Choose n =4 , we have F(2)F(n − 2) + F(3)F(n − 3)
3
2
2
2 3 a +b +c 2 1 2 2 1 3 3 3 3 1
(1) = 0, ab + bc + ca = − 2 = − F(2), abc = (a + b + c )= F(3),F(n)=
3
3
2
5
1
1
n =5 , we have F(2)F(n − 2) + F(3)F(n − 3) (2) = F(2)F(3). Hence,
1
1
(5) = F(2)F(3) + F(3)F
2 2 3 3 6
F(5) F(2) F(3) a + b + c 5 a + b + c 2 a + b + c 3
3
3
5
2
2
5
= · , that is, = .
5 2 3 5 2 · 3
1.121 Given x = by + cz, y = cz + ax, z = ax + by , find the value of
a b c
+ + .
a +1 b +1 c +1
Solution: From the given conditions, we have
x − y = by + cz − cz − ax ⇒ (a + 1)x =(b + 1)y ;
y − z = cz + ax − ax − by ⇒ (b + 1)y =(c + 1)z ;
z − x = ax + by − by − cz ⇒ (c + 1)z =(a + 1)x .
Hence, (a + 1)x =(b + 1)y =(c + 1)z . Let (a + 1)x =(b + 1)y =(c + 1)z = k , then
(ax + by + cz) +(x + y + z) =3k (i). Add up the given equalities in the problem:
x + y + z = 2(ax + by + cz) (ii). (i)(ii) lead to ax + by + cz = k , thus
a b c ax by cz ax + by + cz k
+ + = + + = = =1.
a +1 b +1 c +1 (a + 1)x (b + 1)y (c + 1)z k k
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