Question

Please give answer of the above question

Answer 1

$we \: know \: that \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0 \\ = > {a}^{3} + {b}^{3} + {c}^{3} = 3abc \\ here \: we \: can \: compair \\ {( {a}^{2} - {b}^{2} )}^{3} = {a}^{3} \\ { ({b}^{2} - {c}^{2} ) }^{3} = {b}^{3} \\ {( {c}^{2} - {a}^{2} )}^{3} = {c}^{3} \\ so \\ {( {a}^{2} - {b}^{2} )}^{3} + {( {b}^{2} - {c}^{2}) }^{3} + {( {c}^{2} - {a}^{2} )}^{3} \\ \\ = 3.( {a}^{2} - {b}^{2} ).( {b}^{2} - {c}^{2} ).( {c}^{2} - {a}^{2} ) \\ \\ = 3.(a + b)(a - b)(b + c)(b - c)(c + a)(c - a) \\ \\ next \\ \\ {(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3} \\ \\ = 3.(a - b).(b - c).(c - a) \\ now \: \\ \\ \frac{ {( {a}^{2} - {b}^{2} ) }^{3} + {( {b}^{2} - {c}^{2} ) }^{3} + {( {c}^{2} - {a}^{2} )}^{3} }{ {(a - b)}^{3} + {(b - c)}^{3} } + {(c - a)}^{3} \\ \\ = \frac{3(a + b)(a - b)(b + c)(b - c)(c + a)(c - a)}{3.(a - b)(b - c)(c - a)} \\ = 3.(a + b)(b + c)(c - a)$

the answer. is

3.(a+b)(b+c)(c+a) ✔✔✔✔✔✔