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

Read More:

- Worksheet on Expansion of (a ± b)^2 and its Corollaries
- Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries

### 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