Question

if x + y + z =0 show that x cube + y cube + z cube =3xyz
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Answer 1

Step-by-step explanation:

We have,

x + y + z = 0

Now, we know that,

x³ + y³ + z³ - 3xyz

= (x + y + z)(x² + y² + z² - (xy + yz + xz))

But we are given,

x + y + z = 0

Then,

x³ + y³ + z³ - 3xyz

= (0)(x² + y² + z² - (xy + yz + xz))

Thus,

x³ + y³ + z³ - 3xyz = 0

x³ + y³ + z³ = 3xyz

Hence proved.

Hope it helped you and believing you understood it...All the best

Answer 2

Answer:Given : x+y+z=0

x^3+y^3+z^3=3xyz

the above expression can be written as like

x^3+y^3+z^3-3xyz=

[x+y+z] [x^2+y^2+z^2-xy-yz-zx]

subsititute the given value x+y+z=0 in the above equation

therefore we will get

x^3+y^3+z^3-3xyz=0[x^2+y^2-xy-yz-zx]

if any value is multiplied with zero than total value will be equal to zero

x^3+y^3+z^3-3xyz=0

x^3+y^3+z^3=3xyz

hence proved

Step-by-step explanation: