How Lagrange Interpolation works by Chinese Remainder Theorem Find the polynomial of $2$nd degree such that 
a. $p(1)=0,p(2)=0,p(3)=4$
b.$p(1)=0,p(2)=2,p(3)=0$
c. $p(1)=6,p(2)=0,p(3)=0$ and
d. $p(1)=6,p(2)=2,p(3)=4$.
The solutions for $a,b,c$ are straightforward. For (d) it was given that the sum of three preceding solutions gives the desired polynomial.It is not obvious to me why such is the case.
 A: Let $p_a(x)$ be the polynomial that satisfies the conditions in $a$, and similarly for $p_b(x)$ and $p_c(x)$. You are looking now for $p_d(x)$. 
The book is pointing out that $p_d(1) = p_a(1)+p_b(1)+p_c(1)$. 
So too is $p_d(2) = p_a(2)+p_b(2)+p_c(2)$, and similarly for $p_d(3)$.
So $p_a(x)+p_b(x)+p_c(x)$ meets the conditions of $d$, and you can conclude $p_d(x)=p_a(x)+p_b(x)+p_c(x)$.
A: The exercise is most likely intended to motivate Lagrange interpolation. Here is the basic idea. Let $\,\Bbb P_2\,$ be the polynomials of degree at most $2$. Consider the evaluation map $\,E\,:\ \Bbb P_2\to \Bbb R^3\,$ such that $\, E(p(x)) = [p(1),p(2),p(3)].\,$ Construct basis polynomials $\,p_1,p_2,p_3\,$ such that
$$\begin{array}{rll} p_1 =\ \frac{1}2\!\!\!\!\!&(x-2)(x-3)\overset{E}\longrightarrow  [1,0,0]\\
\\
p_2 =\ &\!\!\!\!\!-(x-1)(x-3)\overset{E}\longrightarrow [0,1,0]\\
\\
p_2 =\ \frac{1}2\!\!\!\!\!&(x-1)(x-2)\overset{E}\longrightarrow  [0,0,1]\end{array}\qquad$$
Note $\,E\,$ is $\,\Bbb R$-linear, i.e. $\,E(r p + s q) = r E(p) + s E(q)\,$ for $\,r,s\in\Bbb R,\,$ therefore
$$ E(r_1 p_1 + r_2 p_2 + r_3 p_3)\, =\, r_1[1,0,0]+r_2[0,1,0]+r_3[0,0,1] \,=\,[r_1,r_2,r_3]$$
Thus $\,E\,$ is surjective. This yields the existence of a solution to the problem of interpolating a quadratic polynomial $\,p(x)\,$ with values $\,r_1,r_2,r_3\,$ at $\,x = 1,2,3.$
Also $\,E\,$ is injective, which yields the uniqueness of the interpolating polynomial. Indeed,  $\,E(p)=E(q)\,\Rightarrow [0,0,0] = E(p)-E(q) = E(p-q).\,$ Therefore $\,p-q = 0,\,$ being a quadratic with $3$ roots $\,x=1,2,3,\,$ i.e. more roots than its degree.
Remark $\ $ The same idea works generally for $\,i=1,\ldots,n.\,$ We seek $\,p_i(x)\,$ ith $\,p_i(i) =1\,$ and $\,p_i(j) = 0\,$ if $\,j\neq i.\,$ Thus all $\,x-j\,$ divide $\,p_i\,$ except  $\,\color{#c00}{x-i}.\,$ Therefore we can write $\,p_i\,$ as follows, where the $\,\rm\color{#c00}{red}$ term is omitted $$\require{cancel}\,p_i(x)\, =\, c_i(x\!-\!1)(x\!-\!2)\cdots\color{#c00}{\cancel{(x-i)}}\cdots (x-n)\,$$
Finally, from $\,p_i(i) = 1\,$ we can evaluate the above at $\,x=i\,$ to solve for $\,c_i.\,$
If you know some linear algebra then you may find it helpful to express the above more precisely in that language (I avoided that in case it is not known). If you know the ring-theoretic Chinese Remainder Theorem (CRT) then you can view such Lagrange interpolation as a special case of applying CRT to the system
$$\begin{align}
p(x) \equiv r_1 \pmod{x-1}\\
p(x) \equiv r_2 \pmod{x-2}\\
p(x) \equiv r_3 \pmod{x-3} \end{align}\qquad$$
