If a+b+c=3, ab+bc+ca=−10 and abc=−24, then the value of
bc
a
2

​
+
ac
b
2

​
+
ab
c
2

​
is:

We need to evaluate:
bc
a
2

​
+
ac
b
2

​
+
ab
c
2

​
=
abc
a
3
+b
3
+c
3

​

Using the algebraic identity:

a
3
+b
3
+c
3
−3abc=(a+b+c)(a
2
+b
2
+c
2
−(ab+bc+ca))
We also know that:

a
2
+b
2
+c
2
=(a+b+c)
2
−2(ab+bc+ca)
Substitute the given values into this expression:

a
2
+b
2
+c
2
=(3)
2
−2(−10)=9+20=29
Now substitute the values into the primary identity:

a
3
+b
3
+c
3
−3(−24)=(3)(29−(−10))
a
3
+b
3
+c
3
+72=3(29+10)
a
3
+b
3
+c
3
+72=3(39)=117
a
3
+b
3
+c
3
=117−72=45
Now, evaluate the target expression:

abc
a
3
+b
3
+c
3

​
=
−24
45
​
=−
8
15
​