(c) (i) Show that $P(x)=x^{4}-3 x^{3}-15 x^{2}-17 x-6$ has a triple zero.

Step 1 ：Let $x$ be a triple zero of $P(x) = x^4 - 3x^3 - 15x^2 - 17x - 6$. Then $P(x) = (x - a)^3(x - b)$ for some integers $a$ and $b$.

Step 2 ：Expand the expression: $(x - a)^3(x - b) = x^4 - (3a + b)x^3 + (3a^2 + 3ab)x^2 - (a^3 + 3a^2b)x + a^3b$

Step 3 ：By comparing coefficients, we have the following system of equations:\\$-3a - b = -3$\\$3a^2 + 3ab = -15$\\$-a^3 - 3a^2b = -17$\\$a^3b = -6$

Step 4 ：From the first equation, we get $b = 3 - 3a$.

Step 5 ：Substitute this into the second equation: $3a^2 + 3a(3 - 3a) = -15$. Simplify to get $9a^2 - 9a - 15 = 0$.

Step 6 ：Factor the quadratic equation: $(3a - 5)(3a + 3) = 0$. So, $a = \frac{5}{3}$ or $a = -1$. Since $a$ must be an integer, we have $a = -1$.

Step 7 ：Substitute $a = -1$ into the equation for $b$: $b = 3 - 3(-1) = 6$.

Step 8 ：Now we have $P(x) = (x + 1)^3(x - 6)$.

Step 9 ：Thus, $P(x)$ has a triple zero at $x = \boxed{-1}.$