Find (x+y)^2/xy +(y+z)^2/yz(zx)^2/zx if x3+y3+z3=3xyz and x+y+z=0

1.	Find (x+y)^2/xy +(y+z)^2/yz(zx)^2/zx if x3+y3+z3=3xyz and x+y+z=0
Answer» Answer $\underbrace{\overbrace{ \orange{ \sf{ \dfrac{ {(<klux>X</klux> + y)}^{2} }{xy} + \dfrac{ {(y + <klux>Z</klux>)}^{2} }{yz} + \dfrac{ {(z + x)}^{2} }{xz} = 3 }}}}$ Given $\red{ \bigstar} \: \sf{x + y + z = 0}$ $\red{ \bigstar} \: \sf{ {x}^{3} + {y}^{3} + {z}^{3} = 3 \: xyz }$ To find $\red{ \bigstar} \: \sf{ \dfrac{ {(x + y)}^{2} }{xy} + \dfrac{ {(y + z)}^{2} }{yz} + \dfrac{ {(z + x)}^{2} }{xz} }$ Solution As GIVEN $\red{\implies }\sf{ {x}^{3} + {y}^{3} + {z}^{3} = 3 \: xyz }$ dividing by xyz both sides $\red{\implies} \sf{ \dfrac{ {x}^{3} }{xyz} + \dfrac{ {y}^{3} }{xyz} + \dfrac{ {z}^{3} }{xyz} = \dfrac{3 \: xyz}{xyz} }$ $\red{\implies} \sf{ \dfrac{ {x}^{2} }{yz} + \dfrac{ {y}^{2} }{xz} + \dfrac{ {z}^{2} }{xy} = 3}$ since, x + y + z = 0 therefore, x = - (y + z) y = - (z + x) z = - (x + y) so, $\red{ \implies} \sf{ \dfrac{ {( -( y + z))}^{2} }{yz} + \dfrac{ {( - (x + z))}^{2} }{xz} + \dfrac{ {( - (x + y))}^{2} }{xy} = 3 }$ $\red{ \implies} \sf{ \dfrac{ {(y + z)}^{2} }{yz} + \dfrac{ {(x + z)}^{2} }{xz} + \dfrac{ {(x + y)}^{2} }{xy} = 3}$ $\boxed{ \boxed{ \red{ \implies \: \sf{ \dfrac{ {(x + y)}^{2} }{xy} + \dfrac{ {(y + z)}^{2} }{yz} + \dfrac{ {(z + x)}^{2} }{xz} = 3 }}}} \huge{\dagger}$ Answer.

Answer»

Answer

$\underbrace{\overbrace{ \orange{ \sf{ \dfrac{ {(<klux>X</klux> + y)}^{2} }{xy} + \dfrac{ {(y + <klux>Z</klux>)}^{2} }{yz} + \dfrac{ {(z + x)}^{2} }{xz} = 3 }}}}$

Given

$\red{ \bigstar} \: \sf{x + y + z = 0}$

$\red{ \bigstar} \: \sf{ {x}^{3} + {y}^{3} + {z}^{3} = 3 \: xyz }$

To find

$\red{ \bigstar} \: \sf{ \dfrac{ {(x + y)}^{2} }{xy} + \dfrac{ {(y + z)}^{2} }{yz} + \dfrac{ {(z + x)}^{2} }{xz} }$

Solution

As GIVEN

$\red{\implies }\sf{ {x}^{3} + {y}^{3} + {z}^{3} = 3 \: xyz }$

dividing by xyz both sides

$\red{\implies} \sf{ \dfrac{ {x}^{3} }{xyz} + \dfrac{ {y}^{3} }{xyz} + \dfrac{ {z}^{3} }{xyz} = \dfrac{3 \: xyz}{xyz} }$

$\red{\implies} \sf{ \dfrac{ {x}^{2} }{yz} + \dfrac{ {y}^{2} }{xz} + \dfrac{ {z}^{2} }{xy} = 3}$

since,

x + y + z = 0

therefore,

x = - (y + z)

y = - (z + x)

z = - (x + y)

so,

$\red{ \implies} \sf{ \dfrac{ {( -( y + z))}^{2} }{yz} + \dfrac{ {( - (x + z))}^{2} }{xz} + \dfrac{ {( - (x + y))}^{2} }{xy} = 3 }$

$\red{ \implies} \sf{ \dfrac{ {(y + z)}^{2} }{yz} + \dfrac{ {(x + z)}^{2} }{xz} + \dfrac{ {(x + y)}^{2} }{xy} = 3}$

$\boxed{ \boxed{ \red{ \implies \: \sf{ \dfrac{ {(x + y)}^{2} }{xy} + \dfrac{ {(y + z)}^{2} }{yz} + \dfrac{ {(z + x)}^{2} }{xz} = 3 }}}} \huge{\dagger}$

Answer.