If x+y+z=4, xy+yz+zx=1, and x
3
+y
3
+z
3
=34, then what is the value of 6xyz?

We know the standard 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:

(x+y+z)
2
=x
2
+y
2
+z
2
+2(xy+yz+zx)
4
2
=x
2
+y
2
+z
2
+2(1)
16=x
2
+y
2
+z
2
+2⟹x
2
+y
2
+z
2
=14
Substitute these values into the main identity:

34−3xyz=4×(14−1)
34−3xyz=4×13
34−3xyz=52
−3xyz=52−34=18
xyz=−6
We need to find the value of 6xyz:

6xyz=6×(−6)=−36