a3 + b3 + c3 − 3abc

MOPSS
28 April 2025

Mathematics Olympiad
Problem Solving Sessions

MOPSS

Department of Mathematics
IISER Bhopal

https://jpsaha.github.io/MOTP/MOPSS/

Suggested readings
• Evan Chen’s advice On reading solutions, available at https://blog.
evanchen.cc/2017/03/06/on-reading-solutions/.
• Evan Chen’s Advice for writing proofs/Remarks on English, available at
https://web.evanchen.cc/handouts/english/english.pdf.
• Notes on proofs by Evan Chen from OTIS Excerpts [Che25, Chapter 1].
• Tips for writing up solutions by Edward Barbeau, available at https:
//www.math.utoronto.ca/barbeau/writingup.pdf.
• Evan Chen discusses why math olympiads are a valuable experience for
high schoolers in the post on Lessons from math olympiads, available at
https://blog.evanchen.cc/2018/01/05/lessons-from-math-olympiads/.
List of problems and examples
1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Example (Moscow MO 1940 Grades 7–8 P1) . . . . . . . . . 3
1.3 Example (India RMO 2002 P2) . . . . . . . . . . . . . . . . 4
1.4 Example (Formula of Unity/The Third Millennium 2022/2023
Qualifying Round Grade R11 P5) . . . . . . . . . . . . . . . 4
1.5 Example (Formula of Unity/The Third Millennium 2023/2024
Qualifying Round Grade R11 P3, S. Pavlov) . . . . . . . . . 5
1.6 Example (India INMO 2002 P2) . . . . . . . . . . . . . . . . 5

§1 a3 + b3 + c3 − 3abc
Example 1.1. Let a, b, c be real numbers. Show that

a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)

= (a + b + c) (a + b + c)2 − 3(ab + bc + ca)

1
= (a + b + c) (a − b)2 + (b − c)2 + (c − a)2 .


2

Remark. An immediate approach would be to begin from the expression

(a + b + c)(a2 + b2 + c2 − ab − bc − ca) at RHS (the right-hand side), multiply
it out and the cancellations would lead to the expression a3 + b3 + c3 − 3abc.
This would definitely provide a proof of the above. However, there is another
way to argue as below.

Solution 1. Observe that

a2 + b2 + c2 − ab − bc − ca
= a2 + b2 + c2 + 2ab + 2bc + 2ca − 3(ab + bc + ca)
= (a + b + c)2 − 3(ab + bc + ca),
2(a2 + b2 + c2 − ab − bc − ca)
= a2 − 2ab + b2 + b2 − 2bc + c2 + c2 − 2ca + a2
= (a − b)2 + (b − c)2 + (c − a)2 .

Note that

a3 + b3 + c3 − 3abc
= (a + b)3 − 3ab(a + b) + c3 − 3abc
= (a + b)3 + c3 − 3ab(a + b) − 3abc
= (a + b)3 + c3 − 3ab(a + b + c)

2
1 a3 + b3 + c3 − 3abc Typos may be reported to jpsaha@iiserb.ac.in.

= (a + b + c)3 − 3(a + b)c(a + b + c) − 3ab(a + b + c)

= (a + b + c)((a + b + c)2 − 3(a + b)c − 3ab)
= (a + b + c)(a2 + b2 + c2 + 2ab + 2bc + 2ca − 3ab − 3bc − 3ca)
= (a + b + c)(a2 + b2 + c2 − ab − bc − ca).
■

Remark. There is another way to prove the above identity.

Solution 2. Consider the polynomial

P (X) = X 3 − (a + b + c)X 2 + (ab + bc + ca)X − abc.
Since a, b, c are the roots1 of the equation P (X) = 0, we obtain
a3 − (a + b + c)a2 + (ab + bc + ca)a − abc = 0,
b3 − (a + b + c)b2 + (ab + bc + ca)b − abc = 0,
c3 − (a + b + c)c2 + (ab + bc + ca)c − abc = 0.
Adding them yields
a3 + b3 + c3 − (a + b + c)(a2 + b2 + c2 ) + (ab + bc + ca)(a + b + c) − 3abc = 0.
This proves that
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca).
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The above identity has the following immediate consequence.

Corollary
If a, b, c are real numbers satisfying a + b + c = 0, then

a3 + b3 + c3 = 3abc.

Example 1.2 (Moscow MO 1940 Grades 7–8 P1). Factor (x − y)3 + (y − z)3 +
(z − x)3 .

Solution 3. Note that if a + b + c = 0, then a3 + b3 + c3 = 3abc. This gives

(x − y)3 + (y − z)3 + (z − x)3 = 3(x − y)(y − z)(z − x).
■

1 If it is not clear, then the following equalities may directly be verified.

Some style files, prepared by Evan Chen, have been adapted here. 3
28 April 2025 https://jpsaha.github.io/MOTP/

Remark. The following proof is direct, and of course, it works.

(x − y)3 + (y − z)3 + (z − x)3

= x3 − 3x2 y + 3xy 2 − y 3
+ y 3 − 3y 2 z + 3yz 2 − z 3
+ z 3 − 3z 2 x + 3zx2 − x3
= −3x2 y + 3xy 2 − 3y 2 z + 3yz 2 − 3z 2 x + 3zx2
= −3xy(x − y) − 3y 2 z + 3yz 2 − 3z 2 x + 3zx2
= −3xy(x − y) − 3y 2 z + 3zx2 + 3yz 2 − 3z 2 x
= −3xy(x − y) + 3z(x2 − y 2 ) − 3z 2 (x − y)
= −3xy(x − y) + 3z(x − y)(x + y) − 3z 2 (x − y)
= 3(x − y) −xy + z(x + y) − z 2

= 3(x − y) −xy + zx + zy − z 2




= 3(x − y) −x(y − z) + z(y − z)
= 3(x − y)(y − z)(z − x).

However, the former solution is less cumbersome, and more elegant.

Example 1.3 (India RMO 2002 P2). Solve the following equation for real x:

(x2 + x − 2)3 + (2x2 − x − 1)3 = 27(x2 − 1)3 .

Solution 4. The given equation is equivalent to

(x2 + x − 2)3 + (2x2 − x − 1)3 + (−3x2 + 3)3 = 0.

Note that x2 + x − 2, 2x2 − x − 1, −3x2 + 3 add up to zero. This implies

(x2 + x − 2)3 + (2x2 − x − 1)3 + (−3x2 + 3)3

= 3(x2 + x − 2)(2x2 − x − 1)(−3x2 + 3)
= −9(x + 2)(x − 1)(x − 1)(2x − 1)(x − 1)(x + 1).

Thus the required solutions for x are

1
−2, −1, , 1.
2
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Example 1.4 (Formula of Unity/The Third Millennium 2022/2023 Qualifying

Round Grade R11 P5). Find all real a, b, c such that
2 2 2
27a +b+c+1
+ 27b +c+a+1
+ 27c +a+b+1
= 3.

4 The content posted here and at this blog by Evan Chen are quite useful.
1 a3 + b3 + c3 − 3abc Typos may be reported to jpsaha@iiserb.ac.in.

Solution 5. For any three real numbers a, b and c, note that

2 2 2
27a +b+c+1
+ 27b +c+a+1
+ 27c +a+b+1

2 2 2
≥ 3 · 3a +b+c+1 · 3b +c+a+1 · 3c +a+b+1
(using Example 1.1 and that 3x ≥ 0 for any real number x)
2
+b2 +c2 +2a+2b+2c+3
= 3 · 3a
2
+(b+1)2 +(c+1)2
= 3 · 3(a+1)
hold. This shows that if a, b, c are real numbers satisfying the given condition,
then
a = b = c = −1.
Moreover, note that for a = b = c = −1, the equality
2 2 2
27a +b+c+1
+ 27b +c+a+1
+ 27c +a+b+1
=3
holds. Hence, the solution of the given equation is
a = b = c = −1.
■

Example 1.5 (Formula of Unity/The Third Millennium 2023/2024 Qualifying

Round Grade R11 P3, S. Pavlov). Let a, b, c be nonzero real numbers such
that
a b c b c a
+ + = 6, + + = 2.
b c a a b c
What could be the value of the expression
a3 b3 c3
+ + ?
b3 c3 a3

Solution 6. Write x = ab , y = cb , z = ac . Note that

x + y + z = 6, xy + yz + zx = 2.
This yields
x3 + y 3 + z 3 = 3 + (x + y + z)(x2 + y 2 + z 2 − xy − yz − zx)
= 3 + (x + y + z) (x + y + z)2 − 3(xy + yz + zx)

= 3 + 6 × (62 − 3 · 2)
= 183.
■

Example 1.6 (India INMO 2002 P2). Find the smallest positive value taken
by a3 + b3 + c3 − 3abc for positive integers a, b, c. Find all a, b, c which give the
smallest value.

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28 April 2025 https://jpsaha.github.io/MOTP/

Walkthrough —
(a) Note that a = b = c = 1 won’t work, not even taking all of a, b, c to be
equal would be of any use. In other words, at least two of a, b, c have to
be unequal.
(b) By taking a = 1, b = 2, c = 1, one can find that a3 + b3 + c3 − 3abc = 4.
Next, we need determine whether a3 + b3 + c3 − 3abc can be equal to
1, 2, 3 or 4 for positive integers a, b, c.
(c) Use

a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)

1
= (a + b + c)((a − b)2 + (b − c)2 + (c − a)2 )
2

to get a lower bound on a3 + b3 + c3 − 3abc.

Solution 7. Let a, b, c be positive integers such that a3 + b3 + c3 − 3abc is

positive. Note that they cannot be equal, and hence at least two of them are
distinct. Since a3 + b3 + c3 − 3abc is symmetric2 in a, b, c, we may assume3
that a ̸= b.
Apart from the integers a and b, there is another pair of two integers among
a, b, c which are not equal, i.e. b ̸= c or c =
̸ a holds. Indeed, if both of these
two inequalities fail to hold, then b = c and c = a hold, and then we would
have a = b, which is a contradiction. Note that
a3 + b3 + c3 − 3abc
= (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
1
= (a + b + c) (a − b)2 + (b − c)2 + (c − a)2


2
1
≥ (a + b + c)(12 + 12 )
2
(since at least two of a − b, b − c, c − a are nonzero, and a + b + c > 0)
≥a+b+c
≥1+2+1 (since at least two of a − b, b − c, c − a are nonzero, and a, b, c ≥ 1)
= 4.
Also note that if c > 1, then
a3 + b3 + c3 − 3abc > 4.
For a = 1, b = 2, c = 1, we obtain
a3 + b3 + c3 − 3abc = 4.
2A reader unfamiliar with this term may require to look online.
3 How we may do so? It does require a thought.

6 The content posted here and at this blog by Evan Chen are quite useful.
References Typos may be reported to jpsaha@iiserb.ac.in.

Hence, the smallest positive value taken by a3 + b3 + c3 − 3abc, for positive

integers a, b, c, is equal to 4.
Moreover, if a, b, c are positive integers such that a3 + b3 + c3 − 3abc takes
the value 4, then at least two of a, b, c are unequal, and the above argument
shows that
a + b + c ≤ a3 + b3 + c3 − 3abc ≤ 4,
and consequently, two of a, b, c are equal to 1 and the remaining one is equal
to 2. Hence, a3 + b3 + c3 − 3abc takes the value 4 precisely when

(a, b, c) = (1, 1, 2), (1, 2, 1), (2, 1, 1).

■
For more exercises around this theme, we refer to [AE11, §1.1].

References
[AE11] Titu Andreescu and Bogdan Enescu. Mathematical Olympiad
treasures. Second. Birkhäuser/Springer, New York, 2011, pp. viii+253.
isbn: 978-0-8176-8252-1; 978-0-8176-8253-8 (cited p. 7)
[Che25] Evan Chen. The OTIS Excerpts. Available at https : / / web .
evanchen.cc/excerpts.html. 2025, pp. vi+289

Some style files, prepared by Evan Chen, have been adapted here. 7