Log in to add a comment

## Answers

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

It is not clear what is to be solved...

x³ + y³ + z³ - 3 x y z = (x + y + z) (x² + y² + z² - xy - y z - zx ) --- (1)

we are given x³ + y³ + z³ = 3 xy z ---(2)

x³+y³ = 3 x y z - z³ = z (3xy - z²)

As LHS = 0, and x+y+z ≠ 0 in (1),

x² + y² + z² = x y + y z + z x --- (3)

(x³ + y³) / z³ = (3 x y z - z³) / z³ = 3 x y /z² - 1 -- (4)

we know that, (x³ + y³) = (x+y) (x² + y² - xy)

So if it is given that (x³ + y³) > z³

=> z ( 3 x y - z²) > z³

=> if z > 0, then 3 x y > 2 z²

if z < 0, then 3 x y < 2 z²

x³ + y³ + z³ - 3 x y z = (x + y + z) (x² + y² + z² - xy - y z - zx ) --- (1)

we are given x³ + y³ + z³ = 3 xy z ---(2)

x³+y³ = 3 x y z - z³ = z (3xy - z²)

As LHS = 0, and x+y+z ≠ 0 in (1),

x² + y² + z² = x y + y z + z x --- (3)

(x³ + y³) / z³ = (3 x y z - z³) / z³ = 3 x y /z² - 1 -- (4)

we know that, (x³ + y³) = (x+y) (x² + y² - xy)

So if it is given that (x³ + y³) > z³

=> z ( 3 x y - z²) > z³

=> if z > 0, then 3 x y > 2 z²

if z < 0, then 3 x y < 2 z²

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

(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0

as x+y+z=+ve real number , x^2+y^2+z^2-xy-yz-zx=0

1/2{(x-y)^2+(y-z)^2+(z-x)^2}=0

as (x-y)^2,(y-z)^2&(z-x)^2 are real and positive number the above equation can

only be possibble when x=y=z

then (x^3+y^3)/z3=2

Hence 2 is the answer