Question

If (x+y+z) = 8 and xy + yz + zx = 20 , then , find x^3 + y^3 + z^3 - 3xyz.

Answer 1

${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)$
${8}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(20)$
${x}^{2} + {y}^{2} + {z}^{2} = 64 - 40$
${x}^{2} + {y }^{2} + {z}^{2} = 24$
now,
${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z )( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$
$= 8(24 - (20))$
$= 8(24 - 20)$
$= 8 \times 4$
$= 32$

Answer 2

Answer:

hope it helps a lot mate....