If $a+b+c=1$, $ab+bc+ca=-8$ and $a^3+b^3+c^3=-11$, then what is the value of $\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}$?

We need to find the value of:
$$X=\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}$$
Taking the common denominator as $abc$, we get:
$$X=\frac{a^2+b^2+c^2}{abc}$$
Step 1: Find $a^2+b^2+c^2$ Using the algebraic identity:
$$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$$
Substitute the given values $a+b+c=1$ and $ab+bc+ca=-8$:
$$(1{)}^2=a^2+b^2+c^2+2(-8)$$
$$1=a^2+b^2+c^2-16$$
$$a^2+b^2+c^2=1+16=17$$
Step 2: Find $abc$ Using another algebraic identity:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))$$
Substitute the known values:
$$-11-3abc=(1)(17-(-8))$$
$$-11-3abc=17+8$$
$$-11-3abc=25$$
$$-3abc=25+11$$
$$-3abc=36$$
$$abc=-12$$
Step 3: Calculate the required value
$$X=\frac{a^2+b^2+c^2}{abc}=\frac{17}{-12}=-\frac{17}{12}$$