# Express a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ - ab - bc - ca as Sum of Squares

Here we will express a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca as sum of squares.

a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca = $$\frac{1}{2}$${2a$$^{2}$$ + 2b$$^{2}$$ + 2c$$^{2}$$ – 2ab – 2bc – 2ca}

= $$\frac{1}{2}$${(a$$^{2}$$ + b$$^{2}$$ – 2ab) + (b$$^{2}$$ + c$$^{2}$$ – 2bc) + (c$$^{2}$$ + a$$^{2}$$ – 2ca)}

= $$\frac{1}{2}$${(a - b)$$^{2}$$ + (b - c)$$^{2}$$ + (c – a)$$^{2}$$}

Corollaries:

(i) If a, b, c are real numbers then (a – b)$$^{2}$$, (b – c)$$^{2}$$ and (c – a)$$^{2}$$ are positive as square of every real number is positive. So,

a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca is always positive.

(ii) a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ – ab – bc – ca = 0 if $$\frac{1}{2}$${(a - b)$$^{2}$$ + (b - c)$$^{2}$$ + (c – a)$$^{2}$$} = 0

Or, (a - b)$$^{2}$$ = 0, (b - c)$$^{2}$$ = 0, (c – a)$$^{2}$$= 0

Or, a - b = 0, b - c = 0, c – a = 0, i.e., a = b = c

Solved Examples on Express a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares:

1. Express 4x$$^{2}$$ + 9y$$^{2}$$  + z$$^{2}$$  – 6xy – 3yz – 2zx as sum of perfect squares.

Solution:

Given expression = 4x$$^{2}$$ + 9y$$^{2}$$ + z$$^{2}$$ – 6xy – 3yz – 2zx

= (2x)$$^{2}$$ + (3y)$$^{2}$$ + z$$^{2}$$ – (2x)(3y) – (3y)(z) – (z)(2x)

= ½[(2x - 3y)$$^{2}$$ + (3y - z)$$^{2}$$ + (z - 2x) $$^{2}$$].

2. If p$$^{2}$$ + 4q$$^{2}$$ + 25r$$^{2}$$ = 2pq + 10qr + 5rp, prove that p = 2q = 5r.

Solution:

Here, p$$^{2}$$ + 4q$$^{2}$$ + 25r$$^{2}$$ = 2pq + 10qr + 5rp

Or, p$$^{2}$$ + 4q$$^{2}$$ + 25r$$^{2}$$ - 2pq - 10qr - 5rp = 0

Or, (p)$$^{2}$$ + (2q)$$^{2}$$ + (5r)$$^{2}$$ – (p)(2q) – (2q)(5r) – (5r)(p) = 0

Or, ½[(p – 2q)$$^{2}$$ + (2q – 5r)$$^{2}$$ + (5r – p)$$^{2}$$] = 0.

If sum of three positive numbers is zero, each number must be equal to 0.

Therefore, p – 2q = 0, 2q – 5r = 0, 5r – p = 0

Thus, p = 2q, 2q = 5r, 5r = p.

Therefore, p = 2q = 5r.

Practice Problems on Express a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ - ab - bc - ca as Sum of Squares:

1. Express each of the following as a sum of perfect squares.

(i) x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$ + xy + yz - zx

[Hint: Given expression = x$$^{2}$$ + (-y)$$^{2}$$ + z$$^{2}$$ - x(-y) -(-y)z - zx

= ½[{x - (-y)}$$^{2}$$ + {(-y) - z}$$^{2}$$ + (z - x)$$^{2}$$.]

(ii) 16a$$^{2}$$ + b$$^{2}$$ + 9c$$^{2}$$ - 4ab - 3bc - 12ca

(iii) a$$^{2}$$ + 25b$$^{2}$$ + 4 - 5ab - 10b - 2a

2. If 4x$$^{2}$$ + 9y$$^{2}$$ + 16z$$^{2}$$ - 6xy - 12yz - 8zx = 0, prove that 2x = 3y = 4z.

3. If a$$^{2}$$ + b$$^{2}$$ + 4c$$^{2}$$ = ab + 2bc + 2ca, prove that a = b = 2c.

1. (i)  ½[(x + y)$$^{2}$$ + (y + z)$$^{2}$$ + (z - x)$$^{2}$$]

(ii) ½[(4a - b)$$^{2}$$ + (b - 3c)$$^{2}$$ + (3c - 4a)$$^{2}$$]

(iii) ½[(a - 5b)$$^{2}$$ + (5b - 2)$$^{2}$$ + (2 - a)$$^{2}$$]

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