Question

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

Answer 1

I would like the answer to this question

Answer 2

Since A has implied an exponent of 1. Simply ad the exponents. AxAxA=A^3
\[a ^{2}\times a ^{1}=a ^{3}\] etc.

Answer 3

I think asker is talking about vectors, not scalars.

Answer 4

lol not that its cross product

Answer 5

no

Answer 6

i am asking cross product in functions

Answer 7

Never mind lol.

Answer 8

AxA = 0 because the A is parallel to itself. Hence AxAxA = 0 also

Answer 9

Make sense?

Answer 10

no i didnt ask that

Answer 11

What's your question then.

Answer 12

i am asking abot ordered pairs

Answer 13

You mean this identity?
\[ A \times B \times C = B(A\cdot C) - C(A \cdot B) \]

Answer 14

\[ A \times (B \times C) \]