Question

if x+y+z=8 and xy+yz+zz+20 find x^3+y^3+z^3-3xyz

Answer 1

here is your answer dear!

Given x + y + z = 8 and xy + yz + zx = 20

Consider, x + y + z = 8

Squaring on both sides, we get

(x + y + z)² = 82

x²+ y² + z² +2(xy + yz +zx) = 64

⇒ x²+ y² + z² = 64 − 2(xy + yz +zx)

= 64 − 2(20) = 64 − 40 = 24

∴ x² + y² + z² = 24

We know that x³+ y³ + z³ − 3xyz = (x + y + z)( x2 + y2 + z2 − xy − yz − zx)

= 8(24 −20) = 8(4) = 32

hope it helps

#be brainly