Question

prove that a^3 + b^3 + c^3 - 3abc = 1/2(a + b + c)[(a-b)^2+(b-c)^2+(c-a)^2]

Answer 1

proved in photo

hope this helps

Answer 2

$<b> <I>$
Hey there !!

→ Prove that :-)

$\bf {a}^{3} + {b}^{3} + {c}^{3} - 3abc = \frac{1}{2} (a + b + c)( {(a - b) }^{2} + {(b - c)}^{2} + {(c - a)}^{2} ).$

→ solution :-)

Using Identity :-

=> a³ + b³ + c³ - 3abc = ( a + b + c )( a² + b² + c² -ab - bc - ca ).

▶ We can write RHS in this form :-

$\bf = \frac{1}{2} (a + b + c)(2 {a}^{2} + 2 {b}^{2} + 2 {c}^{2} - 2ab - 2bc - 2ca).$

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➡ Because , $\bf = \frac{1}{\cancel{2}} (a + b + c) \times \cancel{2} ( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca).$

= ( a + b + c )( a² + b² + c² -ab - bc - ca ).

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$\bf = \frac{1}{2} (a + b + c)( {a}^{2} + {a}^{2} + {b}^{2} + {b}^{2} + {c}^{2} + {c}^{2} - 2ab - 2bc - 2ca).$

= $\bf{ \frac{1}{2} }$ ( a + b + c ) [ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( c² - 2ca + a² ) ].

$\bf { {a}^{3} + {b}^{3} + {c}^{3} - 3abc = \frac{1}{2} (a + b + c)( {(a - b) }^{2} + {(b - c)}^{2} + {(c - a)}^{2} ). }$

✔✔ Hence, it is proved ✅✅.

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