Math, asked by opopop7, 5 months ago

PLEASE WRITE ONLY CORRECT ANSWER...​

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Answered by arwa1910
2

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

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