if a square + b square + c square - A B - B C - CA=0then prove that a = B = C
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Consider,
a^2 + b^2 + c^2 – ab – bc – ca = 0
Multiply both sides with 2, we get....
2( a^2 + b^2 + c^2 – ab – bc – ca) = 0
⇒ 2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0
⇒ (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 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.
________________
#Be Brainly✌️☺️
______________
Consider,
a^2 + b^2 + c^2 – ab – bc – ca = 0
Multiply both sides with 2, we get....
2( a^2 + b^2 + c^2 – ab – bc – ca) = 0
⇒ 2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0
⇒ (a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 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.
________________
#Be Brainly✌️☺️
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