(c) (i) Show that $P(x)=x^4-3x^3-15x^2-17x-6$ 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: $(x-a{)}^3(x-b)=x^4-(3a+b)x^3+(3a^2+3ab)x^2-(a^3+3a^{2}b)x+a^{3}b$

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

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 $x=\boxed{{\displaystyle -1}}.$