If a+b+c=7, and a
2
+b
2
+c
2
=33, then what is the value of a
3
+b
3
+c
3
−3abc?

We know the algebraic identity:
a
3
+b
3
+c
3
−3abc=(a+b+c)(a
2
+b
2
+c
2
−(ab+bc+ca))
First, find the value of (ab+bc+ca) using the identity:

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

(7)
2
=33+2(ab+bc+ca)
49=33+2(ab+bc+ca)
16=2(ab+bc+ca)⟹ab+bc+ca=8
Now, substitute these into the main identity:

a
3
+b
3
+c
3
−3abc=7×(33−8)
a
3
+b
3
+c
3
−3abc=7×25=175