Proof or disprove that the following statement is true I have a problem I need help with:

Prove or disprove that P(M $\cup$ N) $\neq$ P(M) $\cup$ P(N).

What I have tried so  far

Consider: M = $\{ a,b,c\}$ and N = $\{\text{red},\text{blue},\text{yellow}\}$ 

$\Rightarrow$ P(M) ${}= 3^3 = 27$ and P(N) ${} = 3^3$

Combine P(M)$\cup$P(N)

$\Rightarrow \bigcup \{M,N\} = \{x\mid(x\in M) \text{ and } (x\in N)\}$

$\Rightarrow$ P(M ${}\cup{}$ N) ${}= \{a,b,c,\text{red}, \text{blue}, \text{yellow} \}$

$\Rightarrow$ P(M,N) ${}= 6^6 = 46656$

$\Rightarrow$ P(M $\cup$ N) ${}\neq{}$ P(M) ${}\cup{}$ P(N)

Is my answer correct?
Your answer has a good idea but there are some technical mistakes in it. 

For example, you say $P(M)=3^3$, which is not true. 


First of all, $P(M)$ is a set, while $3^3$ is a number. The two are not equal.
What you probably want to say $|P(M)|=3^3$, i.e. that the size of the set $P(M)$ is $3^3$. But even then, no, that is not true.




To try the proof again, I suggest you try to construct a smaller counterexample, and ditch the idea of set sizes. Simply look at the power sets directly.



For an added challenge, try proving this statement:


  $P(A\cup B)=P(A)\cup P(B)$ if and only if $A=\emptyset$ or $B=\emptyset$.
In your example,  $\vert A\vert=\vert B\vert = 3$ and $\vert A \cup B \vert=6$ so 
$\vert P (A)\vert =\vert P (B)\vert =2^3$ and 
$\vert P (A \cup B) \vert =2^6$. ..
Now $\vert P (A) \cup P (B)\vert= \vert P (A)\vert +\vert P (B)\vert - \vert P (A)\cap P (B)\vert=2^3+2^3-1=2^4-1= 15$, 
 so they are not equal. ..