Question

13. If x+y+z=0, show that x + y + z = 3xyz.​

Answer 1

✯✯ QUESTION ✯✯

If x+y+z=0, show that x³ + y³ + z³ = 3xyz.

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✰✰ ANSWER ✰✰

$\longmapsto\tt{x+y+z={x}^{3}+{y}^{3}+{z}^{3}=3xyz}$

➥Using : -

$\longmapsto\tt\bold{{x}^{3}+{y}^{3}+{z}^{3}-3xyz}$

➥Now ,

$\longmapsto\tt{{x}^{3}+{y}^{3}+{z}^{3}-3xyz=(x+y+z)({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx)}$

$\longmapsto\tt{{x}^{3}+{y}^{3}+{z}^{3}=0({x}^{2}+{y}^{2}+{z}^{2}-xy-yz-zx}$

$\longmapsto\tt{{x}^{3}+{y}^{3}+{z}^{3}-3xyz=0}$

$\longmapsto\tt{{x}^{3}+{y}^{3}+{z}^{3}=3xyz}$

✺ HENCE PROVED ✺

Answer 2

Ur Que. is wrong.

It's should be .....

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

We know that,

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

putting x+y+z=0,

x³+y³+z³-3xyz=(0)(x²+y²+z²-xy-yz-zx)

x³+y³+z³-3xyz=0

z³+y³+z³=3xyz

Hence proved.

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