Question

Find the value of x³+ y³ + z³ – 3xyz if x² + y² + z² = 83 and x + y + z = 15​

Answer 1

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