360 problems for mathematical contests

'I'D
ANDREESCD
DORIN
ANDRICA
360
Problems
for
Mathematical
Contests
TITU
ANDREESCU DORIN ANDRICA
360
Problems
for
Mathematical Contests
GIL Publishing House
©
GIL
Publishing
House
ISBN 973-9417-12-4
360 Problems for Mathematical Contests
Authors: Titu Andreescu, Dorin Andrica
Copyright © 2003
by
Gil. All rights reserved.
GIL
Publishing House

P.o.
Box 44, Post Office 3, 4700, Zalau, Romania,
tel.
(+40) 260/616314
fax. (+40) 260/616414
e-mail:
www.gil.ro
IMPRIMERIA
~
ARTA
IV
GRAFICA
I
~~LIBRIS
Calea
$erbanVodti
133,S.4,Cod
70517,BUCURE$TI
Tel.:
336
29
11
Fax:
337
07
35
Contents
FOREWORD

3

FROM
THE
AUTHORS

5
Chapter
1.
ALGEBRA

7
Problems

9
Solutions

17
Chapter
2. NUMBER THEORY

.47
Problems

49
Solutions

57
Chapter
3. GEOMETRY

85

Problems

87
Solutions

95
Chapter
4.
TRIGONOMETRY

137
Problems

139
Solutions

147
Chapter
5.
MATHEMATICAL ANALYSIS

179
Problems

181
Solutions

189
Chapter
6. COMPREHENSIVE PROBLEMS

221
Problems

223
Solutions

235
FOREWORD
I take great pleasure in recommending
to
all readers - Romanians or from abroad
the
book of professors
Titu
Andreescu
and
Dorin Andrica. This book is
the
fruit of a
prodigious activity of
the
two authors, well-known creators of mathematics questions
for Olympiads
and
other mathematical contests. They have published innumerable
original problems in various mathematical journals.
The
book is organized in six chapters: algebra, number theory, geometry,

trigonometry, analysis
and
comprehensive problems. In addition, other fields of math-
ematics found their place in this book, for example, combinatorial problems can be
found in
the
last chapter,
and
problems involving complex numbers are included in
the
trigonometry section. Moreover, in all chapters of this book
the
serious reader can
find numerous challenging inequality problems. All featured problems are interesting,
with
an
increased level of difficulty; some of
them
are real gems
that
will give great
satisfaction
to
any
math
lover attempting
to
solve or even extend them.
Through their outstanding work as
jury

members of
the
National Mathematical
Olympiad,
the
Balkan Mathematics Contest (BMO),
and
the
International Math-
ematical Olympiad
(IMO),
the
authors also supported
the
excellent results of
the
Romanian contestants in these competitions. A great effort was given in preparing
lectures for summer
and
winter training camps
and
also for creating original problems
to
be used in selection tests
to
search for truly gifted mathematics students. To support
the
claim
that
the

Romanian students selected
to
represent
the
country were really
the
ones
to
deserve such honor,
we
note
that
only two mathematicians of Romanian
origin,
both
former IMO gold-medalists, were invited recently
to
give conferences
at
the
International Mathematical Congress: Dan Voiculescu (Zurich, 1994)
and
Daniel
Tataru
(Beijing, 2002).
The
Romanian mathematical community unanimously recog-
nized this outstanding activity of professors
Titu
Andreescu

and
Dorin Andrica. As a
consequence,
Titu
Andreescu,
at
that
time professor
at
Loga Academy in Timi§oara
and
having students on
the
team participating in
the
IMO, was appointed
to
serve
as deputy leader of
the
national team. Nowadays,
Titu's
potential, as with other Ro-
manians in different fields, has been fully realized in
the
United States, leading
the
USA team in
the
IMO, coordinating

the
training
and
selection of team contestants
and
serving as member of several national and regional mathematical contest juries.
One more time, I strongly express my belief
that
the
360 mathematics problems
featured in this book
will reveal
the
beauty of mathematics
to
all students
and
it will
be a guide
to
their teachers
and
professors.
Professor loan Tomescu
Department of Mathematics and Computer Science
University of Bucharest
Associate member of
the
Romanian Academy
FROM

THE
AUTHORS
This book
is
intended
to
help students preparing for all rounds of Mathematical
Olympiads or any other significant mathematics contest. Teachers will also find this
work useful in training young talented students.
Our
experience as contestants was a great asset in preparing this book. To this
we
added our vast personal experience from
the
other side of
the"
barricade" , as creators
of problems
and
members of numerous contest committees.
All
the
featured problems are supposed
to
be original.
They
are
the
fruit of our
collaboration for

the
last 30 years with several elementary mathematics journals from
all over
the
world. Many of these problems were used in contests throughout these
years, from
the
first round
to
the
international level. It
is
possible
that
some problems
are already known,
but
this
is
not critical.
The
important
thing
is
that
an
educated
-
to
a certain extent - reader will find in this book problems

that
bring something
new
and
will teach new ways of dealing with key mathematics concepts, a variety of
methods, tactics,
and
strategies.
The
problems are divided in chapters, although this division
is
not firm, for some
of
the
problems require background in several fields of mathematics.
Besides
the
traditional fields: algebra, geometry, trigonometry
and
analysis,
we
devoted
an
entire chapter
to
number theory, because many contest problems require
knowledge in this field.
The
comprehensive problems in
the

last chapter are also intended
to
help under-
graduate
students
participating in mathematics contests hone their problem solving
skills. Students
and
teachers can find here ideas
and
questions
that
can be interesting
topics for mathematics circles.
Due
to
the
difficulty level of
the
problems contained in this book,
we
deemed
it
appropriate
to
give a very clear
and
complete presentation of all solutions.
In
many

cases, alternative solutions are provided.
As a piece of advice
to
all readers,
we
suggest
that
they
try
to
find
their
own
solutions
to
the
problems before reading
the
given ones. Many problems can be solved
in multiple ways
and
pertain
to
interesting extensions.
This edition
is
significantly different from
the
2002 Romanian edition.
It

features
more recent problems, enhanced solutions, along with references for all published
problems.
We wish
to
extend our gratitude
to
everyone who influenced in one way or another
the
final version of this book.
We
will gladly receive any observation from
the
readers.
The
authors
Chapter
1
ALGEBRA
PROBLEMS
1.
Let C be a set of n characters
{Cl'
C2,
•
, cn}.
We
call
word

a string of
at
most m characters, m
:::;
n,
that
does not
start
nor end with
Cl.
How many
words
can be formed with
the
characters of
the
set
C?
2.
The
numbers
1,2,

, 5n are divided into two disjoint sets. Prove
that
these
sets contain
at
least n pairs (x, y), x > y, such
that

the
number x - y is also
an
element of
the
set which contains
the
pair.
3.
Let
al,
a2,

,an
be distinct numbers from
the
interval
[a,
b]
and let a be a
permutation of
{I,
2,

, n}.
Define
the
function f :
[a,
b]

-t
[a,
b]
as follows:
f(x)
=
{aq(i)
if x =
~i'
i =
1,n
x otherwIse
Prove
that
there is a positive integer h such
that
f[h](X)
fafa

af.
' v "
htimes
x, where f[h]
4.
Prove
that
if
x,
y, z are nonzero real numbers with x + y + z =
0,

then
x
2
+
y2
y2
+
Z2
Z2
+ x
2
x
3
y3
Z3
+
+
=-+-+
x + y y + Z Z + x
yz
zx
xy
5.
Let a,
b,
c,
d be complex numbers with a + b + C + d =
O.
Prove
that

a
3
+ b
3
+ c
3
+ d
3
= 3(abc + bcd + cda + dab).
6.
Let a,
b,
c be nonzero real numbers such
that
a + b + c = 0
and
a
3
+ b
3
+ c
3
=
a
5
+ b
5
+ c
5
•

Prove
that
7.
Let a,
b,
c,
d be integers. Prove
that
a + b + c + d divides
2(a
4
+ b
4
+ c
4
+~)
_ (a
2
+ b
2
+ c
2
+ d
2
)2
+ 8abcd.
10
1.
ALGEBRA
8.

Solve in complex numbers
the
equation
(x
+
l)(x
+
2)(x
+
3)2(X
+
4)(x
+
5)
= 360.
9.
Solve in real numbers
the
equation
-IX
+
VY
+
2v'z=2
+ .jU +
-IV
= x + y + z + u +
v.
10.
Find

the
real solutions
to
the
equation
(x
+
y)
2 =
(X
+
1)
(y
- 1).
11.
Solve
the
equation
~
x+
J4X
+ V
16X
+ l·· +
J4
n
x+3
;x
=
1.

12.
Solve
the
equation
-J
x + a +
-J
x + b +
-J
x + c =
-J
x + a + b -
c,
where
a,
b,
c are real parameters. Discuss
the
equation in terms of
the
values of the
parameters.
13.
Let a
and
b be distinct positive real numbers. Find all pairs of positive real
numbers (x,
y), solutions
to
the

system of equations
14.
Solve
the
equation
{
x4 -
y4
=
ax
-
by
x
2
_
y2
=
{/
a2
_ b
2
.
[
25x -
2] =
l3x
+ 4
4
3'
where

[a]
denotes
the
integer
part
of a real number
a.
.
1+V5
15.
Prove
that
If
a
~
2-'
then
1.1.
PROBLEMS
11
16.
Prove
that
if x,
y,
z are real numbers such
that
x
S
+

yS
+ ZS
f:.
0,
then
the
ratio
2xyz
- (x + y + z)
x
S
+
yS
+ ZS
equals
~
if and only if x + y + z =
O.
17.
Solve in real numbers the equation
1
vx;:-=-r
+ 2.JX2 - 4 +

+
nJXn
- n
2
=
"2(xl

+
X2
+

+ x
n
).
18.
Find
the
real solutions
to
the system of equations
1
11
-+-=9
x y
1 1 1 1
-+-
1+-
1+-
-18
(?Ix
~)(
?Ix)(
~)
-
19.
Solve in real numbers the system of equations
y2

+ u
2
+ v
2
+ w
2
=
4x
- 1
x
2
+ u
2
+ v
2
+ w
2
= 4y - 1
x
2
+
y2
+ v
2
+ w
2
= 4u - 1
x
2
+

y2
+ u
2
+ w
2
= 4v - 1
x
2
+
y2
+ u
2
+ v
2
= 4w - 1
20.
Let
aI,
a2,
as,
a4,
a5
be real numbers such
that
al
+
a2
+ as +
a4
+

a5
= 0 and
max
lai -
ajl
< 1. Prove
that
ai
+
a~
+
a~
+
a~
+
a~
<
10.
l~i<j~5
- -
21.
Let
a,
b,
c be positive real numbers. Prove
that
1 1 1 1 1 1
-+-+-
>
+ +

2a
2b
2c
- a + b b + c c + a
22.
Let
a,
b,
c be real numbers such
that
the sum of any two of
them
is
not
equal
to
zero. Prove
that
23.
Let
a,
b,
c be real numbers such
that
abc = 1. Prove
that
at
most two of the
numbers
are greater

than
1.
1
2a-
-
b'
1
2b-
-
c'
1
2c-
-
a
12
1.
ALGEBRA
24.
Let
a,
b,
c,
d be real numbers. Prove
that
2 2
~
2 1
min(a - b b - c c - a- d - a ) < -
, , , -
4'

25.
Let
aI,
a2,'

,an
be numbers in
the
interval (0,1)
and
let k
;:::
2 be
an
integer
Find
the
maximum value of
the
expression
n
L
V'
a
i(1 -
ai+1),
i=l
26.
Let m
and

n be positive integers. Prove
that
xmn
-1
xn-l
>
m - x
for any positive real number x.
27. Prove
that
m!
;:::
(n!)[~]
for all positive integers m and
n.
28.
Prove
that
1 1 1
~2
1+-+-+'''+->n

v'2 v'3
\Iii n + 1
for any integer n
;:::
2.
29.
Prove
that

n
(1-
1/
vn) + 1 > 1 +
~
+
~
+ +
.!.
> n
(\In
+
1-
1)
2 3 n
for any positive integer n.
( )
na1a2'"
an
30.
Let
aI,
a2,

. , an E 0,1
and
let
tn
= . Prove
that

a1
+
a2
+ + an
n
L loga.
tn
;:::
(n
-
l)n.
i=l
31.
Prove
that
between
nand
3n there is
at
least a perfect cube for any intege
n;:::
10.
32.
Compute
the
sum
n
"'"
1c(Ic+l)
Sn =

~(-1)
2 k.
k=l
33.
Compute
the
sums:
a)
Sn
=
~
(k
+ l)l(k +
2)
(~);
b)
Tn
=
~
(k +
l)(k
+ 2)(k + 3)
(~).
1.1.
PROBLEMS
34.
Show
that
for
any

positive integer n
the
number
Cn;
1)22n
+
Cn:
1)
2
2n
-
2
.3+

+
Cn
2
:
1)3
n
is
the
sum of two consecutive perfect squares.
35.
Evaluate
the
sums:
36.
Prove
that

12
(~)
+ 3
2
(~)
+
52
(~)
+ =
n(n
+ 1)2
n
-
3
for all integers n
~
3.
37.
Prove
that
2
n
L[log2
k]
= (n -
2)2n
+ n + 2
k=l
for all positive integers n.
38.

Let
Xn
= 2
2n
+
1,
n =
1,2,3,

Prove
that
for all positive integers n.
1 2
22
2
n
-
l
1
-+-+-+

+ <-
Xl
X2 X3
Xn
3
39.
Let f : C
-t
C be a function such

that
f(z)f(iz)
=
Z2
for all z E
C.
Prove
that
f (z) + f ( - z) = 0 for all z E C
13
40.
Consider a function f : (0,00)
-t
~
and
a real number a > 0 such
that
f(a) = 1. Prove
that
if
f(x)f(y)
+ f
(~)
f
(~)
=
2f(xy)
for all x, y E (0,00),
then
f is a constant function.

41.
Find with proof if
the
function t:
~
-t
[-1,1],
f(x)
= sin[x]
is
periodical.
n
42.
For all
i,j
=
1,n
define
S(i,j)
=
Lki+i.
Evaluate
the
determinant
~
=
k=l
IS(i,j)l·
14
43.

Let
1.
ALGEBRA
{
ai
if i = j
xii
= 0 if i i
j,
i + j i 2n + 1
b
i
if i + j = 2n + 1
where
ai,
bi
are
real numbers.
Evaluate
the
determinant
~2n
=
IXiil·
44.
a)
Compute
the
determinant
x y z v

y x v z
z v x y
v z y x
b) Prove
that
if
the
numbers
abed,
bade,
cdab,
deba
are divisible by a prime p,
then
at
least
one of
the
numbers
a + b + e +
d,
a + b - e -
d,
a - b + e -
d,
a - b - e +
d,
is divisible
by
p.

45.
Consider
the
quadratic
polynomials
tl
(x) = x
2
+
PI
X +
qr
and
t2
(x) = x
2
+
P2X
+
q~,
where PI,P2, ql,
q2
are
real numbers.
Prove
that
if polynomials
tl
and
t2

have zeros
of
the
same
nature,
then
the
polynomial
has
real zeros.
46.
Let
a,
b,
e
be
real numbers
with
a > 0 such
that
the
quadratic
polynomial
T(x)
=
ax
2
+ bex + b
S
+ e

S
-
4abe
has
nonreal zeros.
Prove
that
exactly one
of
the
polynomials
TI
(x) =
ax
2
+ bx + e
and
T2
(x)
ax
2
+ ex + b
has
only positive values.
47.
Consider
the
polynomials
with
complex coefficients

P(x)
= xn +
alx
n
-
l
+ + an
and
Q(x) = xn +
blx
n
-
l
+ + b
n
having zeros
Xl,
X2,

,X
n
and
xi,
x~,

,x;
respectively.
Prove
that
if al + as +

a5
+ .

and
a2
+
a4
+
a6
+ .

are real numbers,
then
b
l
+ b
2
+ + b
n
is also a real number.
1.1.
BROBLEMS
48.
Let P(x) be a polynomial of degree
n.
If
k
P(k) =
-k-
for k =

0,1,

.
,n
+1
evaluate
P(m),
where m >
n.
49.
Find
all polynomials P(x) with integral coefficients such
that
for all real numbers x.
15
50. Consider
the
polynomials
Pi,
i = 1,
2,

, n with degrees
at
least
1.
Prove
that
if
the

polynomial
P(x)
=
P1(X
n
+
1
)
+
XP2(x
n
+
1
)
+ + x
n
-
1
pn(x
n
+
1
),
is
divisible by x
n
+x
n
-
1

+

·+x+1,
then
all polynomialsPi(x), i =
1,n,
are divisible
by
x-1.
51.
Let P be a prime number
and
let
P(x)
= aoxn + a1xn-1 + +
an
be a polynomial with integral coefficients such
that
an
=1=
0 (mod p). Prove
that
if
there
are n + 1 integers
0'1, 0'2,
•••
, a
n
+ 1 such

that
P ( a
r
)
==
0 (mod
p)
for all
r =
1,2,

,n
+
1,
then
there exist
i,j
with i i- j such
that
ai
==
aj
(mod p).
52. Determine all polynomials P with real coefficients such
that
pn(x)
=
P(x
n
)

for all real numbers x, where n > 1
is
a given integer.
53. Let
P(x)
= aoxn + a1x
n
-
1
+ + an,
an
i- 0,
be a polynomial with complex coefficients such
that
there
is
an
integer m with
Prove
that
the
polynomial P has
at
least a zero with
the
absolute value less
than
1.
54.
Find

all polynomials P of degree n having only real zeros Xl,
X2,

,
Xn
such
that
n 1 n
2
~
P(x)
-
Xi
= XPI(X)'
for all nonzero real numbers x.
55. Consider
the
polynomial with real coefficients
P(x)
= aoxn + a1x
n
-
1
+ + an,
16
1.
ALGEBRA
and
an
f:.

O.
Prove
that
if
the
equation
P(x)
= 0 has all of its roots real
and
distinct,
then
the
equation
x
2
PII(x) +
3xP'(x)
+
P(x)
= 0
has
the
same property.
56.
Let
R~?
and
Rf~)
be
the

sets of polynomials with real coefficients having
no multiple zeros
and
having multiple zeros of order n respectively. Prove
that
if
P(x)
E
R~?
and
P(Q(x)) E
R~?,
then
Q'(x) E
R~]-1).
57. Let
P(x)
be a polynomial with real coefficients of degree
at
least
2.
Prove
that
if there is a real number a such
that
P(a)plI(a) > (P'(a))2,
then
P has
at
least two nonreal zeros.

58. Consider
the
equation
aoxn + a1Xn-1 + +
an
= 0
with real coefficients
ai.
Prove
that
if
the
equation has all of its roots real,
then
(n -
l)ar
2::
2naOa2.
Is
the
reciprocal true?
59.
Solve
the
equation
X4
- (2m + 1)x
3
+ (m - 1)x
2

+
(2m2
+
l)x
+ m = 0,
where m is a real parameter.
60.
Solve
the
equation
x2n
+
a1
x2n
-
1
+ +
a2n_2
X2
- 2nx + 1 = 0,
if all of its roots are real.
SOLUTIONS
1.
Let Nk be
the
number of words having exactly k characters from
the
set
C,
1

:s;
k
:s;
m. Clearly,
N1
= n - 1.
The
number
that
we
seek is
N1
+
N2
+ + N
m
·
Let
Ak
=
{I,
2,

,
k},
1
:s;
k
:s;
m.

We
need
to
find
out
the
number of functions
f : Ak
+
A,
k =
2,
n with
the
properties
f(l)
f:.
a1
and
f(k)
f:.
a1
For
f(l)
and
f(k)
there are n - 1 possibilities of choosing a character from
C2,
•.•
,C

n
and
for
f(i),
1 < i < k there are n such possibilities. Therefore
the
number
of strings
f(l)f(2)

f(k
-
l)f(k)
is
Nk = (n - 1)2n
k
-
2
It
follows
that
N1
+
N2
+ + N
m
= (n -
1)
+ (n -
1)2nO

+ (n - 1)2n
1
+ + (n - 1)2n
m
-
2
=
(Dorin Andrica)
2.
Suppose, for
the
sake of contradiction,
that
there are two sets A
and
B such
that
Au
B =
{I,
2,

,5n},
An
B = 0
and
the
sets contain together less
than
n pairs

(x, y), x > y, with
the
desired property.
Let k be a given number, k
= 1, n.
If
k
and
2k are in
the
same set - A or B -
the
same
can
be said
about
the
difference 2k - k = k.
The
same argument is applied
for
4k
and
2k. Consider
the
case when k
and
4k are elements of A
and
2k

is
an
element of
B.
If
3k is
an
element of
A,
then
4k - 3k = k E
A,
so let 3k E
B.
Now if
5k
E
A,
then
5k -
4k
= k E A
and
if 5k E
B,
then
5k - 3k = 2k E B; so among
the
numbers k, 2k, 3k, 4k, 5k there is
at

least a pair with
the
desired property. Because
k
=
1,2,

,n,
it
follows
that
there are
at
least n pairs with
the
desired property.
(Dorin Andrica, Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 75,
Problem 3698)
3.
Note
that
(1)
17
18
1.
ALGEBRA
and
furthermore
(2)
for all integers

ml,
m2
2::
1.
Suppose
that
for all integers k
2::
1
we
have
I[k](x)
f=.
x.
Because there are n! permutations,
it
follows
that
for k > n! there are distinct
positive integers
nl
> n2 such
that
(3)
Let h =
nl
- n2 > 0
and
observe
that

for all k
the
functions
I[k]
are injective,
since numbers
ai,
i = 1, n are distinct. From relation
(3)
we
derive
that
l[n
2
+h]
(x)
= l[n2]
(x),
x E
[a,
b],
or
(10
l[n2+h-l])(x)
=
(10
l[n
2
-1])(x),
X E

[a,
b].
Because I is injective,
we
obtain
l[n
2
+h-l]
(x)
=
l[n
2
-1]
(x),
x E
[a,
b]
and
in
the
same manner
l[h+l](X)
=
I(x),
x E
[a,
b]
or
l[h](X)
=

x,
x E
[a,
b].
Alternative solution. Let
Sn
be
the
symmetric group of order
nand
Hn
the
cyclic
subgroup generated by
a.
It
is clear
that
Hn
is
a finite group
and
therefore there
is
integer h such
that
a[h]
is
identical permutation.
Notice

that
I[k]
(x)
= {
aq[kJ
(i) if x =
~i'
i = 1, n
x
otherwIse
Then
I[h](x)
= x
and
the
solution is complete.
(Dorin Andrica, Revista Matematica Timi§oara (RMT),
No.
2(1978), pp. 53,
Problem 3540)
4.
Because x + y + z = 0,
we
obtain
x + y =
-z,
y + z =
-x,
z + x =
-y,

or, by squaring
and
rearranging,
x
2
+
y2
=
Z2
_
2xy,
y2
+
Z2
= x
2
_
2yz,
Z2
+ x
2
=
y2
-
2zx.
The
given equality is equivalent
to
Z2
-

2xy
x
2
- 2y Z
y2
-
2zx
x
3
y3
Z3
+
+
=-+-+-,
-z
-x
-y
yz
zx
xy
1.2.
SOLUTIONS
and
consequently
to
(
Xy
yz
zx)
X

3
y3
Z3
-(x+y+z)+2
-+-+-
=-+-+
z x Y
yz
zx
xy
The last equality
is
equivalent
to
2(X
2
y2
+
y2
z
2
+ Z
2
X
2
) =
x4
+
y4
+

Z4.
On
the
other side, from x + y + z = 0
we
obtain (x + y +
Z)2
= 0 or
x
2
+
y2
+
Z2
= -
2(
xy
+ y z +
zx).
Squaring yields
X4
+
y4
+
Z4
+
2(X
2
y2
+

y2
z
2
+ Z
2
X
2
) = 4(X
2
y2
+
y2
z
2
+ Z
2
X
2
) +
8xyz(x
+ y + z)
or
as desired.
19
(Titu
Andreescu, Revista Matematica Timi§oara (RMT) , No. 3(1971), pp. 25,
Problem 483; Gazeta Matematica (GM-B), No. 12(1977), pp. 501, Problem 6090)
5.
We
assume

that
numbers a,
b,
c,
d are different from zero. Consider
the
equation
x4
-
(2:
a)
x
3
+
(2:
ab) x
2
-
(2:
abc) x + abed = 0
with roots a,
b,
c,
d.
Substituting x with a,
b,
e
and
d
and

simplifying by a,
b,
c,
d
f:.
0,
after summing up
we
obtain
Because
2:
a =
0,
it
follows
that
If
one of
the
numbers
is
zero, say
a,
then
b+c+d=
0,
or b + c =
-d.
It
is

left
to
prove
that
b
3
+ c
3
+ d
3
=
3bcd.
Now
b
3
+ c
3
+ d
3
= b
3
+ c
3
-
(b
+
c)3
= -3bc(b +
c)
= 3bcd

as desired.
(Dorin Andrica, Revista Matematica Timi§oara (RMT), No.
1-2(1979), pp. 47,
Problem 3803)
6.
Because a + b + c =
0,
we
obtain
a
3
+ b
3
+ c
3
= 3abc
and
a
5
+ b
5
+ c
5
= 5abc(a
2
+ b
2
+ c
2
+

ab
+
bc
+ ca)