Understanding example 22 of duality from Hoffman and Kunze's Linear Algebra I encountered one example in Hoffman and Kunze's Linear Algebra. 
I did not get why one step happens, so I couldn't proceed further. 


I do not get from where the highlighted portion came. 
Any help will be appreciated.
 A: $L = 0$ means that for any polynomial $f \in V$ we have: $$L(f) = c_1L_1(f)+L_2(f)+c_3L_3(f) = c_1f(t_1)+c_2f(t_2)+c_3f(t_3)=0$$
In particular setting $f(x) = 1$ gives $c_1+c_2+c_3 =0$, $f(x)=x$ gives $c_1t_1+c_2t_2+c_3t_3 =0$ and $f(x) = x^2$ gives $c_1t_1^2+c_2t_2^2+c_3t_3^2 =0$.
A: We want to check the following: given the real vector space $V = \{ p : \mathbb{R} \to \mathbb{R} : p \text{ is a polynomial function} \}$ and real numbers $t_1$, $t_2$ and $t_3$, the linear functionals $L_i : V \to \mathbb{R}$ defined by
$$
L_i(p) = p(t_i), \qquad i = 1,2,3,
$$
are linearly independent. So, we want to check that if the linear functional $L$ given by
$$
L = c_1 L_1 + c_2 L_2 + c_3 L_3
$$
is the zero functional, then each $c_i$ must be zero. Now, $L = 0$ means that $L(p) = 0$ for any $p \in V$. In particular, $L(p_i)$ must be zero for each of the polynomial functions $p_1(x) = 1$, $p_2(x) = x$ and $p_3(x) = x^2$. Evaluating $L$ at each of these polynomial functions, we get the highlighted equations:
\begin{align}
L(p_1) &= c_1L_1(p_1) + c_2 L_2(p_1) + c_3L_3(p_1)\\
&= c_1 p_1(t_1) + c_2 p_1(t_2) + c_3 p_1(t_3) \\
&= c_1 + c_2 + c_3.\\
& \\
L(p_2) &= c_1L_1(p_2) + c_2 L_2(p_2) + c_3L_3(p_2)\\
&= c_1 p_2(t_1) + c_2 p_2(t_2) + c_3 p_2(t_3) \\
&= c_1 t_1 + c_2 t_2 + c_3 t_3.\\
& \\
L(p_3) &= c_1L_1(p_3) + c_2 L_2(p_3) + c_3L_3(p_3)\\
&= c_1 p_3(t_1) + c_2 p_3(t_2) + c_3 p_3(t_3) \\
&= c_1 t_1^2 + c_2 t_2^2 + c_3 t_3^2.\\
\end{align}
So, we have the highlighted equations,
\begin{alignat}{10}
c_1 & {}+{} & c_2 & {}+{} & c_3 & {}={} 0 \\
t_1 c_1 & {}+{} & t_2 c_2 & {}+{} & t_3 c_3 & {}={} 0 \\
t_1^2 c_1 & {}+{}&  t_2^2 c_2 & {}+{} & t_3^2 c_3 & {}={} 0.
\end{alignat}
