a2+b2+c2-ab-bc-ac = 0 prove that a=b=c
Answers
Step-by-step explanation:
a² + b² + c² = ab + bc + ca
On multiplying both sides by “2”, it becomes
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
=> Since the sum of square is zero then each term should be zero
⇒ (a –b)² = 0, (b – c)² = 0, (c – a)² = 0
⇒ (a –b) = 0, (b – c) = 0, (c – a) = 0
⇒ a = b, b = c, c = a
∴ a = b = c.
Step-by-step explanation:
a²+b⅔+c²-ab-bc-ac = 0
(APPLYING TO "2" BOTH SIDE)
2a²+2b²+2c²-2ab-2bc-2ac = 0
SPLITING
AND FOLLOWING
a²-2ab+ b²
(a²-2ab +b²)+ (b²-2bc+ c²)+(c²-2ac+ a²)=0
(a-b)²+(b-c)²+(c-a)² = 0
(a-b)²= 0 (b-c)²= 0
a-b= 0. b-c= 0
a=b b= c
similarly a= c
THEREFORE A=B= C
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