Question

If a^2 + b^2 + c^2-ab-bc-ca = 0 , prove that a=b=c

Answer 1

a^2 + b^2 + c^2-ab-bc-ca = 0 

or,2a^2 + 2b^2 + 2c^2 -2ab -2bc -2ac = 0x2

or,a^2 -2ab +b^2 + a^2 -2ac +c^2 + b^2 -2bc +c^2 =0

or,(a-b)^2+(b-c)^2+(c-a)^2=0

Hence,a=b=c(proved)

Answer 2

Given a² + b² + c² = ab + bc + ca

Multiplying both sides with "2", we have
2 ( a² + b² + c² ) = 2 ( ab + bc + ca)
2a² + 2b² + 2c² = 2ab + 2bc + 2ca
a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca = 0
a² + b² - 2ab + b² + c² - 2bc + c² + a² - 2ca = 0
(a² + b² - 2ab) + (b² + c² - 2bc) + (c² + a² - 2ca) = 0
(a - b)² + (b - c)² + (c - a)² = 0
(a - b)² = (b - c)² = (c - a)² = 0
(a - b)² = 0 ---------- (1)
(b - c)² = 0 ---------- (2)
(c - a)² = 0 ---------- (3)

Simplifying Equ. (1), we have
(a - b)² = 0

Taking Square Root on both sides, we have
a - b = 0
a = b ---------- (4)

Simplifying Equ. (2), we have
(b - c)² = 0

Taking Square Root on both sides, we have
b - c = 0
b = c ---------- (5)

Simplifying Equ. (3), we have
(c - a)² = 0

Taking Square Root on both sides, we have
c - a = 0
c = a ---------- (6)

From Equation No. (4), (5) & (6) , it is proved that
a = b = c

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