Question

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Answer 1

Question:-

• If ${a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca = 0$
• Prove that a = b = c

Solution:-

Given,

${a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca = 0$

Then,

${a}^{2} + {b}^{2} + {c}^{2} = ab + bc + ca$

Multiply both sides with 2

$2( {a}^{2} + {b}^{2} + {c}^{2} ) = 2(ab + bc + ca)$

$2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} = 2ab + 2bc + 2ac$

${a}^{2} + {a}^{2} + {b}^{2} + {b}^{2} + {c}^{2} + {c}^{2} - 2ab - 2bc - 2ac = 0$

$( {a}^{2} + {b}^{2} - 2ab) + ( {b}^{2} + {c}^{2} - 2bc) + ( {c}^{2} + {a}^{2} - 2ac) = 0$

${(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}$

------> Since the sum of square is Zero then each term should equal to Zero

${(a - b)}^{2} = 0 = > a = b$

${(b - c)}^{2} = 0 = > b = c$

${(c - a)}^{2} = 0 = > c = a$

So,

$a = b = c$

Hence, Proved

Answer 2

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