 The formula of a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares is read as a square plus b square plus c square minus ab minus bc minus ca. By substituting this formula the students can solve the problems in Expansion of Powers of Binomials and Trinomials. So, go through this article to know more about how to express a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares and its corollaries. Also, we have presented some examples of expressing a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares at the end of this page.

Express a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares

a² + b² + c² – ab – bc – ca = 1/2{2a² + 2b² + 2c² – 2ab – 2bc – 2ca}
a² + b² + c² – ab – bc – ca = 1/2{(a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca)}
a² + b² + c² – ab – bc – ca = 1/2{(a – b)² + (b – c)² + (c – a)²}

Corollaries:

1. If a, b, c are the real numbers then (a – b)², (b – c)², and (c – a)² are positive.
2. a² + b² + c² – ab – bc – ca is always positive.
3. a² + b² + c² – ab – bc – ca = 0
If 1/2{(a – b)² + (b – c)² + (c – a)²} = 0
(a – b)² = 0
(b – c)² = 0
(c – a)² = 0
or
a – b = 0, b – c = 0, c – a = 0
Thus a = b = c = 0

Expressing a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares Examples

Example 1.
Simplify the equation ( x + y + z) ( x² + 2y² + 3z² – 2xy -3yz – 4zx)
Solution:
Given that the equation is
( x + y + z) ( x² + 2y² + 3z² – 2xy -3yz – 4zx)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (x + y + z) {(x² + y² + z² – 2(x)(y) – 3(y)(z) – 4(z)(x)} = x³ + y³ + z³ – 3xyz

Example 2.
Simplify the equation (2x + y + 3z) (x² + y² + z² + 4z² – 2yz – 2zx)
Solution:
Given that the equation is
(2x + y + 3z) (x² + y² + z² + 4z² – 2yz – 2zx)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (2x + y + 3z){(2x)² + y² + (3z)² + 4(3z)² – 2(y)(3z) – 2(3z)(2x) = (2x)³ + (y)³ + (3z)³ – 3(2x)(y)(3z)
= 8x³ + y³ + 9z³ – 18 xyz
Thus the simplification of (2x + y + 3z) (x² + y² + z² + 4z² – 2yz – 2zx) is 8x³ + y³ + 9z³ – 18 xyz

Example 3.
If a + b + c = 7 and ab + bc + ca = -6 then the value of a³ + b³ + c³ – 3abc.
Solution:
Given that
a + b + c = 7
ab + bc + ca = -6
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
= (7)(a² + b² + c² + 6)
Therefore a³ + b³ + c³ – 3abc = 7a² + 7b² + 7c² + 42

Example 4.
Simplify the equation (a + 3b – 2c)( a² + b² + c² – 2ab + 3b + 8ca)
Solution:
Given that the equation is
(a + 3b – 2c)( a² + b² + c² – 2ab + 3b + 8ca)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (a + 3b + (-2c)){ a² + (3b)² + (-2c)² – 2a(3b) + 3(3b) + 8(-2c)(a)
= a³ + (3b)³ + (-2c)³ – 3(a)(3b)(-2c)
= a³ + 27b – 8c + 18abc
Thus the simplification of (a + 3b – 2c)( a² + b² + c² – 2ab + 3b + 8ca) is a³ + 27b – 8c + 18abc

Example 5.
If a + b + c = 10 and ab + bc + ca = 5 then the value of a³ + b³ + c³ – 3abc
Solution:
Given that
a + b + c = 10
ab + bc + ca = 5
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
= (10)(a² + b² + c² – 5)
Therefore a³ + b³ + c³ – 3abc = 10a² + 10b² + 10c² – 50.

Example 6.
Simplify the equation (x + y + 3z) ( x² + 2y² + z² – 3xy – 4yz – 4zx)
Solution:
Given that the equation is
(x + y + 3z) ( x² + 2y² + z² – 3xy – 4yz – 4zx)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is ( x + y + 3z) ( x² + y² + 3z² – (x)(y) – (y)(3z) – (3z)(x) = x³ + y³ + (3z)³ – 3(x)(y)(3z)
= x³ + y³ + 27z³ – 9xyz
Thus the simplification of (x + y + 3z) ( x² + 2y² + z² – 3xy – 4yz – 4zx) is x³ + y³ + 27z³ – 9xyz

Example 7.
If a + b + c = 2 and ab + bc + ca = 1 then the value of a³ + b³ + c³ – 3abc.
Solution:
Given that
a + b + c = 2
ab + bc + ca = 2
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
= (2)(a² + b² + c² – 1)
Therefore a³ + b³ + c³ – 3abc = 2a² + 2b² + 2c² – 2

Example 8.
Simplify the equation (a + 2b + 2c) ( a² + b² + c² – 2ab + 2bc + 2ca)
Solution:
Given that the equation is
(a + 2b + 2c) (a² + b² + c² – 2ab + 2bc + 2ca)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (a + 2b + 2c) {(a² + (2b)² + (2c)² – (a)(2b) – (2b)(2c) – (2c)(a)} = a³ + (2b)³ + (2c)³ – 3a(2b)(2c)
= a³ + 8b³ + 8c³ – 12abc
Thus the simplification of (a + 2b + 2c) ( a² + b² + c² – 2ab + 2bc + 2ca) is a³ + 8b³ + 8c³ – 12abc