# If x+y+z =positive real number, x^3+ y^3+ z^3=3xyz then solve x^3+ y^3/z^3

2
by pratikkumar399

2015-05-07T23:28:03+05:30

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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

2015-05-08T09:58:55+05:30
Given that x^3+y^3+z^3=3xyz
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