Question

if x+y+z=0 then the square of the value of (x+y)^2/xy+(y+z)^2/yz+(z+x)^2/zx is of

Answer 1

Answer:

(x + y)²/xy  + (y + z)²/yz  + (z + x)²/zx = 3

Step-by-step explanation:

(x + y)²/xy  + (y + z)²/yz  + (z + x)²/zx

= (z(x + y)²)  + x(y + z)² + y²(z + x)²)/xyz

x + y + z =0

=> x + y = -z

y + z = -z

z + x = -y

=> ( z(-z)² + x(-x)² + y(-y)²)/xyz

= ( z³ + x³ + y³)/xyz

= (x³ + y³ + z³)/xyz

as we know that

if x + y + z =  0

then x³ + y³ + z³ = 3xyz

= 3xyz/xyz

= 3

(x + y)²/xy  + (y + z)²/yz  + (z + x)²/zx = 3

Answer 2

Answer:

(x + y)²/xy + (y + z)²/yz + (z + x)²/zx = 3

Step-by-step explanation:

(x + y)²/xy + (y + z)²/yz + (z + x)²/zx

= (z(x + y)²) + x(y + z)² + y²(z + x)²)/xyz

x + y + z =0

=> x + y = -z

y + z = -z

z + x = -y

=> ( z(-z)² + x(-x)² + y(-y)²)/xyz

= ( z³ + x³ + y³)/xyz

= (x³ + y³ + z³)/xyz

as we know that

if x + y + z = 0

then x³ + y³ + z³ = 3xyz

= 3xyz/xyz

= 3

(x + y)²/xy + (y + z)²/yz + (z + x)²/zx = 3