simplify the following
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Solution :-
We need to evaluate
- a² ( b² - c² ) + b² ( c² - a² ) + c² ( a² - b² )
Simplifying ,
⇒ a² ( b² - c² ) + b² ( c² - a² ) + c² ( a² - b² )
⇒ a²b² - a²c² + b²c² - b²a² + c²a² - c²b²
⇒ a²b² - a²b² - a²c² + a²c² + b²c² - b²c²
⇒ 0 - 0 - 0
⇒ 0
Hence , by evaluating a² ( b² - c² ) + b² ( c² - a² ) + c² ( a² - b² ) = 0
More information :-
Algebraic identities ,
- ( a + b )( a - b ) = a² - b²
- ( a + b )² = a² + b² + 2ab
- ( a - b )² = a² + b² - 2ab
- a² + b² = ( a + b )² - 2ab
- ( a + b )³ = a³ + b³ + 3ab ( a + b )
- ( a + b )³ = a³ + b³ + 3a²b + 3ab²
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