If $A=\{-1,1\},$ find $A \times A \times A.$
If is known that for any non-empty set $A, A \times A \times A$ is defined as
$A \times A \times A=\{(a, b, c): a, b, c \in A\}$
It is given that $A=\{-1,1\}$
$\therefore A \times A \times A=\left\{\begin{array}{l}(-1-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1), \\ (1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\end{array}\right\}$
If $P=\{1,2\},$ form the set $P \times P \times P$
If $A \times B =\{(p, q),(p, r),(m, q),(m, r)\},$ find $A$ and $B$
The solution set of $8x \equiv 6(\bmod 14),\,x \in Z$, are
Let $A = \{1, 2, 3, 4, 5\}; B = \{2, 3, 6, 7\}$. Then the number of elements in $(A × B) \cap (B × A)$ is
If $A = \{ x:{x^2} - 5x + 6 = 0\} ,\,B = \{ 2,\,4\} ,\,C = \{ 4,\,5\} ,$ then $A \times (B \cap C)$ is