How does the strong form of the PCP theorem imply the inapproximability of Max-XORSAT?
Theorem 11.4 states that for every $\delta &gt; 0$ there is a polytime computable function that on input $\varphi$ outputs a PCP (of a certain form) such that (i) if $\varphi$ is satisfiable then the cost of the PCP (to be defined) is at least $1-\delta$, and (ii) if $\varphi$ is unsatisfiable then the cost of the PCP is at most $1/2+\delta$.

The PCP consists of a distribution over quadruples $(i,j,k,b)$ where $i,j,k$ are indices in the range $1,\ldots,N$, for some $N$ depending polynomially on the size of $\varphi$, and $b \in \{0,1\}$. The value of an assignment $\alpha\colon \{1,\ldots,N\} \to \{0,1\}$ is equal to $\Pr[\alpha(i) \oplus \alpha(j) \oplus \alpha(k) = b]$, and the value of the PCP is the maximum value of an assignment.

(The version in which $b = 0$ always is easy to solve – the zero assignment has value 1. Similarly, when $b = 1$ the one assignment has value 1.)

Furthermore, the distribution is generated using $t = O(\log n)$ random bits. We can think of $i,j,k,b$ as functions of the randomness $r$.

Consider now the MAX-XORSAT instance
$$
\bigwedge_{r \in \{0,1\}^t} (x_{i(r)} \oplus x_{j(r)} \oplus x_{k(r)} = b(r)).
$$
Since $|\{0,1\}^t| = 2^t = 2^{O(\log n)} = \mathit{poly}(n)$, this instance has size polynomial in the size of $\varphi$.

The maximum fraction of constraints satisfied by this MAX-XORSAT instance is exactly the same as the value of the PCP. Hence if $\varphi$ is satisfiable, at least a $1-\delta$ fraction of constraints can be satisfied, whereas if $\varphi$ is unsatisfiable, at most a $1/2+\delta$ fraction can be satisfied.

An approximation algorithm for MAX-XORSAT with ratio strictly better than $1/2$ can distinguish between the two cases (for small enough $\delta$), and so can find out whether $\varphi$ is satisfiable or not. In other words, such an algorithm can be used to solve SAT.