Question

If x+y+z=6 and xy+yz+zx = 12 , then show that x3+y3+z3 = 3xyz

Answer 1

Given:
1)
$x + y + z = 6 \\ xy + yz + zx = 12$

2) We know that,
${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx) \\ = > {6}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2 \times 12 \\ = > {x}^{2} + { y }^{2} + {z}^{2} = 36 - 24 = 12$

3) And,
We also know that,
${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z) \\ \times ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ = > 6 \times (12 - 12) = 0 \\ = > {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 0 \\ = > {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$
Hence Proved.

Answer 2

Step-by-step explanation:

it is the correct answer ❤️