identity used is
x³ + y³+ z³– 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx).
then use
(x+y+z)² = x²+y²+z²+2(xy+yz+xz)
225= 83 + 2(xy+yz+xz)
xy+yz+xz = (225-83)/2
xy+yz+xz= 142/2
xy+yz+xz= 71
ok
now use identity
x³ + y³+ z³– 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx).
now
x³ + y³+ z³– 3xyz = 15 (83 – xy – yz – zx).
= 15[83 - (71)]
= 15×12
=180
