Question

PLEASE WRITE ONLY CORRECT ANSWER...​

Answer 1

12.

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = \frac{1}{2} (a + b + c)( {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} \\ lhs - = \frac{1}{2} (a + b + c)( {(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2} \\ = \frac{1}{2} (a + b + c)( {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} - 2ab - 2bc - 2ca) \\ = \frac{1}{2} \times 2(a + b + c)( {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} - ab - bc - ca) \\ = 1(a + b + c)( {a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} - ab - bc - ca) \\ = {a}^{3} + {b}^{3 } + {c}^{3} - 3abc \\ lhs = rhs \\ hence \: proved$