(d) For $(a, b), (c, d)$ $\epsilon$  $N × N$

$(a,\,b)R(c,\,d) \Rightarrow ad(b + c) = bc(a + d)$

Reflexive : Since $ab(b + a)$ = $ba(a + b)\forall ab \in N$,

$\therefore $$(a,\,b)R(a,\,b)$, $\therefore $ $R$ is reflexive.

Symmetric : For $(a,\,b),\,(c,\,d) \in N \times N$, let $(a,\,b)R(c,\,d)$

$\therefore $ $ad(b + c) = bc(a + d)$ ==> $bc(a + d) = ad(b + c)$

==> $cb(d + a) = da(c + b)$ ==> $(c,\,d)R(a,\,b)$

$\therefore $ $R$ is symmetric

Transitive : For $(a,\,b),(c,\,d),\,(e,\,f) \in $N$ \times N,$

Let $(a,\,b)R(c,\,d),\,(c,\,d)R(e,\,f)$

$\therefore $ $ad(b + c) = bc(a + d)$, $cf(d + e) = de(c + f)$

$⇒$ $adb + adc = bca + bcd$…..$(i)$

and $cfd + cfe = dec + def$ …..$(ii)$

$(i) ×ef +  (ii) × ab$ gives,

$adbef + adcef + cfdab + cfeab$

= $bcaef + bcdef + decab + defab$

$⇒$ $adcf(b + e) = bcde(a + f)$

$⇒$ $af(b + e) = be(a + f)$

$⇒$ $(a,\,b)R\,(e,\,f)$.

$\therefore $ $R$ is transitive.

Hence $R$ is an equivalence relation.