If a2 - b2 - c2 - ab - bc - ca - 89 and a - b - b - c - c - a - 12 - 5 - 3- where a- b and c are real numbers- then a2 - b2 - c2 - ab - bc - ca equals- a2 - b2 - c2 - ab - bc - ca - 89 - a - b - b - c - c - a - 12 - 5 - 3- - a- b - c - - - - a2 - b2 - c2 - ab - bc - ca - -

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A.M Greater than or Equal to G.M

If a2 + b2 + ...

Question

If a² + b² + c² + ab + bc + ca = 89 and (a + b) : (b + c) : (c + a) = 12 : 5 : 3, where a, b and c are real numbers, then (a² + b² + c² – ab – bc – ca) equals

यदि a² + b² + c² + ab + bc + ca = 89 तथा (a + b) : (b + c) : (c + a) = 12 : 5 : 3, जहाँ a, b व c वास्तविक संख्याएँ हैं, तब (a² + b² + c² – ab – bc – ca) बराबर है

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a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

a^{3} + b^{3} + c^{3} - 3 a b c

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

a + b + c = 12

a^{2} + b^{2} + c^{2} = 50

If (a+b+c) = 12 and a $^{2}$ +b $^{2}$ +c $^{2}$ = 108, then (ab+bc+ca) = ?

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