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

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

Mathematics
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I would like the answer to this question
Since A has implied an exponent of 1. Simply ad the exponents. AxAxA=A^3 \[a ^{2}\times a ^{1}=a ^{3}\] etc.
I think asker is talking about vectors, not scalars.

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Other answers:

lol not that its cross product
no
i am asking cross product in functions
Never mind lol.
AxA = 0 because the A is parallel to itself. Hence AxAxA = 0 also
Make sense?
no i didnt ask that
What's your question then.
i am asking abot ordered pairs
You mean this identity? \[ A \times B \times C = B(A\cdot C) - C(A \cdot B) \]
\[ A \times (B \times C) \]

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