If x+y+z=3 and xy+yz+zx=−11, then what is the value of x
3
+y
3
+z
3
−3xyz?

We use the well-known algebraic identity:
x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−(xy+yz+zx))
First, find x
2
+y
2
+z
2
using the identity:

(x+y+z)
2
=x
2
+y
2
+z
2
+2(xy+yz+zx)
Substitute the given values:

(3)
2
=x
2
+y
2
+z
2
+2(−11)
9=x
2
+y
2
+z
2
−22
x
2
+y
2
+z
2
=9+22=31
Now, substitute these values into the primary identity:

x
3
+y
3
+z
3
−3xyz=3×(31−(−11))
x
3
+y
3
+z
3
−3xyz=3×(31+11)
x
3
+y
3
+z
3
−3xyz=3×42=126