Question

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Answer 1

$use \: \: \boxed{if \: \: a + b + c = 0 \: \:then, \:{a}^{3} + {b}^{3} + {c}^{3} = 3abc }$

Step-by-step explanation:

$Given, \: \: \: a + b + c = 0 \\ Then, \: \: \: b + c = - a, \: \: c + a = - b, \: and \: \: a + b = - c \\ Now, \: \: \: {a}^{2} (b + c) + {b}^{2} (c + a) + {c}^{2} (a + b) + 3abc \\ putting \: \: values \\ = {a}^{2} \times - a + {b}^{2} \times - b + {c}^{2} \times - c + 3abc \\ = - {a}^{3} - {b}^{3} - {c}^{3} - 3abc \\ = - ( {a}^{3} + {b}^{3} + {c}^{3}) + 3abc \\ we \: \: know \: \: that \: \: \boxed{if \: \: a + b + c = 0 \: \:then, \:{a}^{3} + {b}^{3} + {c}^{3} = 3abc } \\ so, \: \: \: = - 3abc + 3abc = 0$

Answer 2

Answer:

In order to prove that

4a^2-b^2+c^2+4ac=0

4a^2+c^2+4ac=b^2

(2a+c)^2=b^2

2a+c=b

2a+c-b=0

From this we get eq1=eq2

Hence proved.