We verify b) and c) (De Morgan's laws) using a) (double-negation law).
a) $\lnot (\lnot P) \leftrightarrow P$.
b) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q)$
then use c) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \land \lnot Q)$ [ rewrite it as : $\lnot [\lnot (\lnot P) \lor \lnot (\lnot Q) ]$ ; now it is of the "form" : $\lnot [\lnot P_1 \lor \lnot Q_1]$; then you must replace $\lnot P_1 \lor \lnot Q_1$ with $\lnot (P_1 \land Q_1)$, by c), that is really : $\lnot (\lnot P \land \lnot Q)$]. In this way you will get :
$\lnot (P \lor Q) \leftrightarrow \lnot (\lnot \lnot P \lor \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \land \lnot Q)$
then apply again double-negation to the right-hand side ("cancelling" $\lnot \lnot$) and you will have :
$\lnot (P \lor Q) \leftrightarrow (\lnot P \land \lnot Q)$.
c) - Start with the left-hand side and put $\lnot \lnot P$ in place of $P$ and $\lnot \lnot Q$ in place of $Q$ (i.e., use double-negation a)) :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q)$
then use b) to transform the content of right-hand side parentheses into : $\lnot (\lnot P \lor \lnot Q)$ getting :
$\lnot (P \land Q) \leftrightarrow \lnot (\lnot \lnot P \land \lnot \lnot Q) \leftrightarrow \lnot \lnot (\lnot P \lor \lnot Q)$
then apply again double-negation and it's done.
The Concensus Theorem says:
$XY \lor Y'Z \lor XZ = XY \lor XZ$
So your exercise is really a particular application of this!
But let's assume the Concensus Theorem is not available for you to use.
Then you can do:
$(A \land B) \lor (\neg A \land C) \lor (B \land C) =$ (Adjacency)
$(A \land B) \lor (\neg A \land C) \lor (A \land B \land C) \lor (\neg A \land B \land C)=$ (Absorption x 2)
$(A \land B) \lor (\neg A \land C)$
This assumes:
Adjacency
$P = (P \land Q) \lor (P \land \neg Q)$
Absorption
$P = P \lor (P \land Q)$
Best Answer
Sorry to disappoint, but if you want to "algebraically" prove the equivalence, you are going to need to use distributivity and DeMorgan's, and a couple more identities.
Since each side of the equivalence has the term $\;(\lnot y \land \lnot z) \lor,\;$
we can first focus on:
$$x \land ((\lnot y \land z) \lor (y \land \lnot z))) \equiv x\land (\lnot (y \land z)))$$
And now you see each side has an equivalent conjunct: $x \land\;$
Indeed, to prove: $$(\lnot y \land \lnot z) \lor (x \land (\lnot y \land z) \lor (y \land \lnot z))) \equiv (\lnot y \land \lnot z) \lor (x \land (\lnot (y \land z)))$$
we first focus on the following to check for equivalence:
$$\lnot (y \land z) \equiv^? (\lnot y \land z) \lor (y \land \lnot z)$$
There is no getting around using the distributive properties, absorption, and DeMorgan's, etc. but this breaks it down so as to make such applications relatively transparent.
$$\lnot (y \land z)\; \equiv^? \;(\lnot y \land z) \lor (y \land \lnot z)$$ $$\equiv\; [(\lnot y \land z) \lor y] \land [(\lnot y \land z )\lor \lnot z]\tag{distributivity}$$ $$\equiv [(\lnot y \lor y) \land (z \lor y)] \land [(\lnot y \lor \lnot z) \land (z \lor \lnot z)]\tag{?}$$
$$ \equiv T \land (z \lor y) \land (\lnot y \lor \lnot z) \land T\tag{?}$$
$$\equiv (z \lor y) \land (\lnot y \lor \lnot z)\tag{?}$$
$$ \vdots$$ $$ \vdots$$
Don't forget that each side of the original equivalence is also also true whenever $\lnot y \land \lnot z \equiv \lnot (y \lor z)$ is true, as it is a disjunct to the expressions on each side, so if it holds, the equivalence holds.
So, e.g., if $x$ is false, then the expressions on the RHS and LHS are equivalent because then, the truth values of each expression is equivalent to the truth value of $\lnot p \land \lnot q$: i.e., each side evaluates to the equivalent truth value.
To prove algebraically, just keep the "disjunct" and $x \land$ "riding along" (operating only as you would above) until you can safely eliminate $x$ and/or make use of $\lnot p \land \lnot q$ to establish the equivalence.
Final Note Using the "$=$"-sign isn't really appropriate in logic. I have used the symbol for logical equivalence: "$\equiv$", which means (equivalently) $\iff$: "if and only if". It can be viewed as a logical connective in its own right. I'll include the truth-table displaying its truth-functional definition: