Question

question is given above in the picture

Answer 1

See 24 it will help

Answer 2

Hope u like my process..
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Formula to be required :
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(x +y)³ = x³ + y³ +3xy(x +y)
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Given :
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a + b + c =0

or, a + b = - c

Cubing both sides we get,
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(a + b)³ = (-c) ³

or, a³ + b³ +3ab(a +b) = - c³

or, a³ + b³ + 3ab(-c) +c³ = 0

or, a³ + b³ + c³ - 3abc = 0

or, a³ + b³ + c³ = 3abc

Dividing both sides by abc
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$\frac{( {a}^{3} + {b}^{3} + {c}^{3}) }{abc} = \frac{3abc}{abc} \\ \\ or. \: \: \frac{ {a}^{3} }{abc} + \frac{ {b}^{3} }{abc} + \frac{ {c}^{3} }{abc} = 3 \\ \\ or. \: \: \frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ac} + \frac{ {c}^{2} }{ab} = 3...proved...$
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