Question

if x+y+z=10 and xy+yz+zx=31 than what is the value of (x+y)² (y+z)² (z+x)²​

Answer 1

Answer:

          138

Step-by-step explanation:

x + y + z = 10

(x + y + z)² = (10)²

x² + y² + z² + 2xy + 2yz + 2zx = 100

x² + y² + z² + 2(xy + yz + zx) = 100                { xy+yz+zx=31}

x² + y² + z² + 2(31) = 100

x² + y² + z² + 62 = 100

x² + y² + z² = 100 - 62

x² + y² + z² = 38 -------{i}

(x+y)² = x² + y² + 2xy

(y+z)² = z² + y² + 2zy

(z+x)² = z² + x² + 2zx

(x+y)² + (y+z)² + (z+x)²​

= (x² + y² + 2xy) + (z² + y² + 2zy) + (z² + x² + 2zx)

= x² + y² + z² + y² + z² + x² + 2xy + 2zy + 2zx

= 2x² + 2y² + 2z²

=  2 (x² + y² + z² + xy + zy + zx)  -------{ii}

from {i} and {ii}

= 2 (x² + y² + z² + xy + zy + zx)                 { xy+yz+zx=31}

= 2 (38 + 31)

= 2 (69)

= 138

∴(x+y)² + (y+z)² + (z+x)²​ = 138