Question

Let's see if you can solve this maths prob​

Answer 1

Answer:

see, it can be solved using

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

$\frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ca} + \frac{ {c}^{2} }{ab} \\ = \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} \\ = {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + \\ c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ = {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0 \times {a}^{2} + \\ {b}^{2} + {c}^{2} - ab - bc - ca \\ = {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0 \\ = {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

putting the values of

${a}^{3} + { b}^{3} + {c}^{3} \\ = \frac{3abc}{abc} \\ = 3$

so mate I solved this problem.

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