(a) To show that R is an equivalence relation, we need to prove that R is reflexive, symmetric, and transitive.

(i) Reflexive:

We need to show that $(a, b) R (a, b)$ for all $(a, b) \in N \times N$.

According to the definition of R, we need to check if $ab(b+a) = ba(a+b)$.

Since $ab(b+a) = ab(a+b) = ba(a+b)$, R is reflexive.

(ii) Symmetric:

We need to show that if $(a, b) R (c, d)$, then $(c, d) R (a, b)$.

Given $(a, b) R (c, d)$, we have $ad(b+c) = bc(a+d)$.

We need to show that $cb(d+a) = da(c+b)$.

Since $ad(b+c) = bc(a+d)$, we can rewrite this as $bc(a+d) = ad(b+c)$.

Rearranging, we have $cb(a+d) = da(b+c)$, which is the same as $cb(d+a) = da(c+b)$.

Thus, $(c, d) R (a, b)$, and R is symmetric.

(iii) Transitive:

We need to show that if $(a, b) R (c, d)$ and $(c, d) R (e, f)$, then $(a, b) R (e, f)$.

Given $(a, b) R (c, d)$, we have $ad(b+c) = bc(a+d)$.

Given $(c, d) R (e, f)$, we have $cf(d+e) = de(c+f)$.

We need to show that $(a, b) R (e, f)$, which means we need to show that $af(b+e) = be(a+f)$.

From $ad(b+c) = bc(a+d)$, we have $\frac{a+d}{ad} = \frac{b+c}{bc}$, which simplifies to $\frac{1}{d} + \frac{1}{a} = \frac{1}{c} + \frac{1}{b}$.

From $cf(d+e) = de(c+f)$, we have $\frac{c+f}{cf} = \frac{d+e}{de}$, which simplifies to $\frac{1}{f} + \frac{1}{c} = \frac{1}{e} + \frac{1}{d}$.

Adding these two equations, we get $\frac{1}{d} + \frac{1}{a} + \frac{1}{f} + \frac{1}{c} = \frac{1}{c} + \frac{1}{b} + \frac{1}{e} + \frac{1}{d}$.

Simplifying, we get $\frac{1}{a} + \frac{1}{f} = \frac{1}{b} + \frac{1}{e}$, which means $\frac{a+f}{af} = \frac{b+e}{be}$.

Therefore, $be(a+f) = af(b+e)$, which means $(a, b) R (e, f)$.

Thus, R is transitive.

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.