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Let $P(x) = Q(x) (x-a)(x-b)(x-c) + R(x)$

Since the degree of divisor is $3$ . So, the degree of R(x) is atmost $2$ .

So, Let $R(x) = px^{2} + qx +r$ .

So, $P(a) = pa^2 + qa + r$ , $P(b) = pb^2 + qb + r$ and $P(c) = pc^2 + qc + r$ .

But, $P(a) = a, P(b) = b, P(c) =c$

So, $a = pa^2 + qa + r$ , $b = pb^2 + qb + r$ and $c = pc^2 + qc + r$ .

It implies that $a-b = p(a^2 - b^2) +q(a-b)$ and $b-c = p(b^2 - c^2) +q(b-c)$

So, $p(a+b) +q = 1$ and $p(b+c) +q = 1$ .

Subtracting the above equations,

We get $p(a-c) = 0$ . But $a \neq c$ .

So, $p = 0$

So, it will imply that $q = 1$ and $r=0$ ( But How? Try yourselves )

So, $\boxed{R(x) = x}$ .

Let $P(x)=Q(x)(x-a)(x-b)(x-c)+R(x)$

Since the degree of the divisor is $3.$ So, the degree of $R(x)$ is $2.$

Let $R(x)$ be $lx^2+mx+n.$

Now, $P(x)=Q(x)(x-a)(x-b)(x-c)+(lx^2+mx+n).$

Given that

$P(a)=a \Rightarrow a=la^2+ma+n$

$P(b)=b \Rightarrow b=lb^2+mb+n$

$P(c)=c \Rightarrow c=lc^2+ml+n$

$(a-b)=la^2-lb^2+ma-mb \Rightarrow (a-b)=l(a+b)(a-b)+m(a-b)$

$\Rightarrow l(a+b)+m =1 ............................ ... \boxed{1}$

Similarly , $(b-c)=lb^2-lc^2+mb-mc \Rightarrow (b-c)=l(b-c)(b+c)+m(b-c)$

$\Rightarrow l(b+c)+m = 1 ...............................\boxed{2}$

Similarly, $(a-c)=la^2-lc^2+ma-mc \Rightarrow (a-c)=l(a-c)(a+c+m(a-c))$

$\Rightarrow l(a+c)+m =1 ...............................\boxed{3}$

Solving $\boxed{1} , \boxed{2}$ ,we get,

$l(a+b) -l(b+c)+m-m = 0.$

$la+lb-lb-lc=0.$

$l(a-c)=0.$

But, $a \neq c \Rightarrow \boxed{ l=0}$

Substituting $l=0$ in $\boxed{1},$ we get ,

$0(a+c)+m=1 \Rightarrow \boxed{m=1}$

Substituting $l,m$ in $P(a),$ we get ,

$a=0(a^2)+1(a)+n \Rightarrow \boxed{n=0.}$

Now, $R(x)=lx^2+mx+n = 0(x^2)+1(x)+0 = \boxed{x}$

Therefore the remainder is $x.$