If \[\text{A=}\left\{ 1,2,3 \right\},\text{ B=}\left\{ 4 \right\},\text{ C=}\left\{ 5 \right\}\] then, verify that \[\text{A}\times \left( \text{B-C} \right)=\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)\]

Hint: To solve this question, we will use these formulas. The Cartesian product of two sets P and Q is given by,
\[\text{P}\times \text{Q=}\left\{ \left( p,q \right) \right\}:p\in P,q\in Q\]
And subtraction of two sets P and Q is given by,
\[P-Q=\left\{ x:x\in P\text{ and x}\notin \text{Q} \right\}\]

Complete step-by-step answer:
Given that, \[\text{A=}\left\{ 1,2,3 \right\},\text{ B=}\left\{ 4 \right\},\text{ C=}\left\{ 5 \right\}\]
We will separately calculate \[\left( \text{A}\times \text{B} \right),\left( \text{A}\times C \right)\text{ and }\left( \text{B-C} \right)\] first. The Cartesian product of two sets P and Q is given by;
\[\text{P}\times \text{Q=}\left\{ \left( p,q \right) \right\}:p\in P,q\in Q\]
And subtraction of two sets P and Q is given by,
\[P-Q=\left\{ x:x\in P\text{ and x}\notin \text{Q} \right\}\]
The possible combination of $A\times B$ using the above formula is given by;
\[A\times B=\left\{ \left( 1,4 \right),\left( 2,4 \right),\left( 3,4 \right) \right\}\]
The possible combination of $A\times C$ using the above formula is given by;
\[A\times C=\left\{ \left( 1,5 \right),\left( 2,5 \right),\left( 3,5 \right) \right\}\]
The possible value of B-C using the above formula of difference of two set is given by;
\[B-C=\varnothing \text{ (empty set)}\]
Now, the value of $A\times \left( B-C \right)$ using the above formula is $\varnothing $ as there is no element in B-C. So, $A\times \left( B-C \right)$ is also an empty set.
Now, consider \[\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)\]
Using the formula of difference of set stated above, we get:
\[\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)=\left\{ \left( 1,4 \right),\left( 2,4 \right),\left( 3,4 \right) \right\}-\left\{ \left( 1,5 \right),\left( 2,5 \right)\left( 3,5 \right) \right\}=\varnothing \]
As there is no common element. Hence, the value of \[\text{A}\times \left( \text{B-C} \right)=\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)=\varnothing \]
Hence verified.

Note: The biggest possibility of mistake which the students make while solving this question is considering \[\text{A}\times \left( \text{B-C} \right)=\left\{ \left( 1,0 \right),\left( 2,0 \right),\left( 3,0 \right) \right\}\] This step is wrong, because 0 and $\varnothing $ are different elements, here \[B-C=\varnothing \text{ (empty set)}\] and not 0. Therefore, \[\text{A}\times \left( \text{B-C} \right)=\left\{ \left( 1,0 \right),\left( 2,0 \right),\left( 3,0 \right) \right\}\] is wrong as \[0\notin B-C\]