How to show that $\bigl(A(x) \land B(x) \bigr)\land C(x) \equiv A(x) \land \bigl(B(x)\land C(x) \bigr)?$ Suppose $A(x),B(x),C(x)$ are statements. (For the purpose of my question, we don't need to specify what exactly they are)
How to show that $\bigl(A(x) \land B(x) \bigr)\land C(x)$  is equivalent to $A(x) \land \bigl(B(x)\land C(x)  \bigr)?$
Surefire way to do so would be to generate truth table, yet I would like to find more concise way to prove that.
This is what I've tried:

Let's denote statement $\bigl(A(x) \land B(x)
 \bigr)\land C(x)$ as $(1)$ and statement $A(x) \land \bigl(B(x)\land
 C(x)  \bigr)$ as $(2)$
Required to prove that $(1) \iff (2)$
$(\rightarrow)$ Let x be some value such that $(1)$ is true and $(2)$ is
  false. Since $(1)$ is true, all statements must be true, but for $(2)$ to be false, at least one statement must be false. Hence if
   $(1)$ is true, then $(2)$ is true.
$(\leftarrow)$ Suppose $(2)$ is true. Suppose $(1)$ is false. Since $(2)$ is true, it follows that all statements are true, but for $(1)$ to be false, at least one statement must be false, hence contradiction. In other words, if $(2)$ is true, then $(1)$ is true.
Therefore, $(1) \iff (2)$ and thus $(1) \equiv (2)$. $\Box$ 

Is it correct? Are there better (and shorter) alternatives to prove the conjecture above? 
 A: Due to various metaproperties of propositional logic together with, say, this natural deduction calculus, we can equivalently show the logical equivalence by showing derivability in both directions.
Formally:

Corollary: We have $(A(x) ∧ B(x)) ∧ C(x) ≡ A(x) ∧ (B(x) ∧ C(x))$ iff.
  
  
*
  
*$(A(x) ∧ B(x)) ∧ C(x) \vdash A(x) ∧ (B(x) ∧ C(x))$
  
*and $A(x) ∧ (B(x) ∧ C(x)) \vdash (A(x) ∧ B(x)) ∧ C(x)$.
  
  
  where $\equiv$ denotes logical equivalence with standard propositional logic and $\vdash$ denotes the natural deduction calculus I linked above.

Proof: I think completeness and soundness play important roles, but would need to look up precise definitions to be really sure.


Lemma: 1. and 2. from above are in fact fulfilled.

Proof: For 1. we apply the introduction rule for conjunction and therefore have to show $A(x)$ and $B(x) \wedge C(x)$ separately. Applying the same rule for the latter conjunction we get in total three subgoals: $A(x)$, $B(x)$ and $C(x)$. We can easily extract them from our hypothesis (left of $\vdash$) by conjunction elimination.
Step 2. works similarly.

As a comment to your solution and the other answer: Personally, I developed the habit of always preferring constructive proofs over proofs involving proof-by-contradiction. Especially with proving with a calculus as done above, we can directly see where the evidence for our proof goals comes from!
A: Assume $$A\land(B\land C)\tag{$P$}$$ and, for a contradiction, $$\lnot((A\land B)\land C).\tag{$Q$}$$
From $(P)$, we have $A$ and $B\land C$, the latter of which gives $B$ and $C$; now, from $(Q)$, we have either $\lnot(A\land B)$ or $\lnot C$, but since $C$, we must have the former, which, in turn, gives either $\lnot A$ or $\lnot B$, but both of which are impossible.
Thus $(P)\implies \lnot (Q)$.
Can you proceed from here?
A: Since you want a concise proof, note that false-is-$0$, true-is-$1$ reduces $\land$ to multiplication, which is famously associative.
