Given $a b c t \neq 0$, satisfying
$$
\left\{\begin{array}{l}
a=t b+c, \\
b=c\left(1+t+t^{2}\right) .
\end{array}\right.
$$
(1) Prove that $a$ is a root of the quadratic equation
$$
c x^{2}+c(b-2 c) x-(b-c)\left(b^{2}+c^{2}\right)=0
$$

When both real roots of the equation are $a$, find the value of $t$.
(2) When $a=15, b=7$, find the values of $c$ and $t$.
(1) It is known that $a, b, c, t$ are all non-zero, so we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
t=\frac{a-c}{b}, \\
b=c\left(1+t+t^{2}\right)
\end{array}\right. \\
\Rightarrow b=c\left[1+\frac{a-c}{b}+\left(\frac{a-c}{b}\right)^{2}\right] \\
\Rightarrow b^{3}=b^{2} c+b c(a-c)+c\left(a^{2}-2 a c+c^{2}\right) \\
\Rightarrow c a^{2}+b c a-2 c^{2} a+b^{2} c-b c^{2}+c^{3}-b^{3}=0
\end{array}
$$

Since $c \neq 0$, $a$ is a root of the quadratic equation
$$
c x^{2}+c(b-2 c) x-(b-c)\left(b^{2}+c^{2}\right)=0
$$

When the equation has a repeated root $a$,
$$
\begin{array}{l}
\Delta=c^{2}(b-2 c)^{2}+4 c(b-c)\left(b^{2}+c^{2}\right) \\
=b^{2} c(4 b-3 c)=0 .
\end{array}
$$

But $b c \neq 0$, so we get $4 b=3 c$, substituting into equation (2) gives
$$
3=4\left(1+t+t^{2}\right) \Leftrightarrow(2 t+1)^{2}=0 \text {. }
$$

Solving this, we get $t=-\frac{1}{2}$.
(2) Substituting $a=15, b=7$ into equation (3) gives
$$
c^{3}-37 c^{2}+379 c-343=0 \text {. }
$$

Since the sum of the coefficients is zero, there must be a solution $c=1$, then
$$
(c-1)\left(c^{2}-36 c+343\right)=0 \text {, }
$$

i.e., $(c-1)\left[(c-18)^{2}+19\right]=0$.
Therefore, $c=1$ is the only solution.
Substituting into equation (1) gives $t=\frac{a-c}{b}=\frac{15-1}{7}=2$.