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AravindG
 4 years ago
is there any easy way to findcross product AxAxA rather than finding product AxA and multi[lying by A??
AravindG
 4 years ago
is there any easy way to findcross product AxAxA rather than finding product AxA and multi[lying by A??

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2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0I would like the answer to this question

radar
 4 years ago
Best ResponseYou've already chosen the best response.0Since A has implied an exponent of 1. Simply ad the exponents. AxAxA=A^3 \[a ^{2}\times a ^{1}=a ^{3}\] etc.

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0I think asker is talking about vectors, not scalars.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1lol not that its cross product

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1i am asking cross product in functions

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0AxA = 0 because the A is parallel to itself. Hence AxAxA = 0 also

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0What's your question then.

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.1i am asking abot ordered pairs

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0You mean this identity? \[ A \times B \times C = B(A\cdot C)  C(A \cdot B) \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.0\[ A \times (B \times C) \]
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