$(p \rightarrow q) \rightarrow r\equiv(\sim p\vee q)\rightarrow r$
$(p \rightarrow q) \rightarrow r\equiv\sim(\sim p\vee q)\vee r$
$(p \rightarrow q) \rightarrow r\equiv(p\wedge \sim q)\vee r$
Now we can change propositional operator into boolean operator for easy calculation
$\wedge\equiv\cdot,\vee\equiv +$
$(p \rightarrow q) \rightarrow r\equiv p. \bar q+ r$ --------------$>(1)$
and $p \rightarrow (q \rightarrow r)\equiv p\rightarrow (\sim q\vee r)$
$p \rightarrow (q \rightarrow r)\equiv \sim p\vee (\sim q\vee r)$
Now we can change propositional operator into boolean operator for easy calculation
$\wedge\equiv\cdot,\vee\equiv +$
$p \rightarrow (q \rightarrow r)\equiv \bar p+ (\bar q+ r)$
$p \rightarrow (q \rightarrow r)\equiv \bar p+ \bar q+ r$ ------------$>(2)$
From equation $(1)$ and $(2)$ both are not
equivalent.