85.If x2 + xy +xz = 135, ŃĐł + yz + xy = 351 and Z2 + ZX + Zy = 2

1.	85.If x2 + xy +xz = 135, ŃĐł + yz + xy = 351 and Z2 + ZX + Zy = 243. Then the value of x is. (x, y, z > 0)
Answer» Given :x^2+xy + xz = 135 --- ( 1 ) , y^2+yz + yx = 351 --- ( 2 ) And z^2+zx + zy = 243 --- ( 3 ) Now we add equation 1 , 2 and 3 and get x^2+y^2 +z^2 + 2xy+ 2yz + 2zx = 729 (x+y+z)^2 = 729 ( We know : (x+y+z)^2 =x^2+y^2 +z^2 + 2xy+ 2yz + 2zx) (x+y+z)^2 = 27^2 x+y+z = 27 --- ( 4 ) Now we take equation 1 :x^2+xy + xz = 135 x(x+y+z) = 135 , Substitute value from equation 4 and get 27x = 135 x= 5 And we take equation 2 :y^2+yz + yx = 351 y(x+y+z) = 351 , Substitute value from equation 4 and get 27y = 351 y= 13 And we take equation 3 :z^2+zx + zy = 243 z(x+y+z) = 243 , Substitute value from equation 4 and get 27z = 243 z= 9 Therefore, Value ofx^2+y^2+z^2 = 5^2 + 13^2+ 9^2= 25 + 169 + 81 =275 thanks

85.If x2 + xy +xz = 135, ŃĐł + yz + xy = 351 and Z2 + ZX + Zy = 243. Then the value of x is. (x, y, z > 0)

Answer»

Given :x^2+xy + xz = 135 --- ( 1 ) ,

y^2+yz + yx = 351 --- ( 2 )

And

z^2+zx + zy = 243 --- ( 3 )

Now we add equation 1 , 2 and 3 and get

x^2+y^2 +z^2 + 2xy+ 2yz + 2zx = 729

(x+y+z)^2 = 729 ( We know : (x+y+z)^2 =x^2+y^2 +z^2 + 2xy+ 2yz + 2zx)

(x+y+z)^2 = 27^2

x+y+z = 27 --- ( 4 )

Now we take equation 1 :x^2+xy + xz = 135

x(x+y+z) = 135 , Substitute value from equation 4 and get

27x = 135

x= 5

And we take equation 2 :y^2+yz + yx = 351

y(x+y+z) = 351 , Substitute value from equation 4 and get

27y = 351

y= 13

And we take equation 3 :z^2+zx + zy = 243

z(x+y+z) = 243 , Substitute value from equation 4 and get

27z = 243

z= 9

Therefore,

Value ofx^2+y^2+z^2 = 5^2 + 13^2+ 9^2= 25 + 169 + 81 =275

thanks