Prove That \((A\cup B)\times C=(A\times C)\cup(B\times C)\)

Let \[ (x,y)\in (A\cup B)\times C \]

Then \[ x\in A\cup B \quad \text{and} \quad y\in C \]

So, \[ x\in A \quad \text{or} \quad x\in B \]

If \[ x\in A \] and \[ y\in C, \] then \[ (x,y)\in A\times C \]

If \[ x\in B \] and \[ y\in C, \] then \[ (x,y)\in B\times C \]

Hence,

\[ (x,y)\in (A\times C)\cup(B\times C) \]

Now let \[ (x,y)\in (A\times C)\cup(B\times C) \]

Then \[ (x,y)\in A\times C \] or \[ (x,y)\in B\times C \]

Therefore,

\[ x\in A \text{ or } x\in B, \quad y\in C \]

Thus,

\[ x\in A\cup B \]

and

\[ (x,y)\in (A\cup B)\times C \]

Hence,

\[ \boxed{ (A\cup B)\times C=(A\times C)\cup(B\times C) } \]