is this language context-free? a tricky one Is this language context-free?

$$L = \{a^nb^nc^{2n} \mid n \ge 0\}$$

It's tricky in my opinion because I know that $a^nb^nc^n$ is not context-free, but can I determine from this that $L$ is not context-free, too?

Thanks in advance

EDIT:
I can use these closures: all the closures of regular languages, the closure of intersection of regular language with a context-free language, and closure under homorphishms
It is not context-free. You can show this using the pumping lemma for context-free languages; the proof is very similar to the one for the language $\{a^nb^nc^n:n&gt;0\}$, which is given in the linked article.

Added: Since you’re restricted to closure properties, perhaps the easiest argument is to use closure under inverse homomorphisms, using the homomorphism $h$ such that $h(a)=a$, $h(b)=b$, and $h(c)=cc$. If $L$ were context-free, $\{a^nb^nc^n:n\ge 0\}$ would also be context-free.
$$L=\{a^nb^nc^2n/n\ge1\}$$
is not context free but it is context sensitive language.
here we can write like below  and think it is cfl 
$a^n b^n c^n c^n$
no of $a$'s plus no of $b$'s is equal or not to no of $c$'s of twice of $a$'s and b's
it should be possible bcz we can maintain a count between no of $b$'s and $c$'s then $a$'s and $c$'s but it is not correct .
reason is we are maintaining count between no of $a$'s plus no of $b$'s but not no of $a$'s equal to no of $b$'s hence it is cfl.