Question

If x+y+z=0 show that x³+y³+z³=3xyz
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Answer 1

Answer:

= Put x + y + z = 0 ,

= x³ + y³ + z³ - 3xyz = ( 0 ) ( x² + y² + z² - xy - yz - zx )

= x³ + y³ + z³ - 3xyz = 0

= x³ + y³ + z³ = 3xyz

Answer 2

Answer:

$x + y + z = 0 \\ to \: prove - {x}^{3} + {y}^{3} + {z}^{3} = 3xyz \\ we \: can \: write \: the \: above \: equation \: as \\ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz \\ = (x + y + z)( {x}^{2} + {y}^{2} + {z }^{2} - xy - yz - zx) \\ = (0)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ = 0 \\ so \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 0 \\ hence \: {x}^{3} + {y}^{3} + {z}^{3} = 3xyz \\ hence \: proved$

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