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If a = 5 . 3 , b = 1 . 1 , c = − 6 . 4 , Evaluate a 3 + b 3 + c 3 − 3 a b c .
The answer is 0.
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3 solutions
Bonus question: Is the following equation true?
a 4 + b 4 + c 4 + d 4 − 4 a b c d = ( a + b + c + d ) ( a 3 + b 3 + c 3 + d 3 − a b c − a b d − a c d − b c d )
Hi! :3 :3 :# → :3*
if a+b+c=0 then a^3+b^3+c^3=3abc
If a+b+c=0, then a^3+b^3+c^3=3abc And according to qustn a+b+c=0 therefore using above eqn answer is 0
a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a )
Here, a + b + c = 0
Hence, RHS=0
Answer 0