Question

if x^3+y^3+z^3=49 and x+y+z=1 then find the value of xy+yz+zx-xyz​

Answer 1

Answer:

If x, y, z are real numbers if x ³ + y ³ + Z ³ = 13, X + Y + Z = 1 and x×y×z =1 what is the value of XY + YZ + ZX?

x3+y3+z3−3xyz

=(x+y+z)(x2+y2+z2−xy−yz−zx)

=(x+y+z)[(x+y+z)2−3(xy+yz+zx)]

∴13−3×1=1×[12−3(xy+yz+zx)]

Solve it and you will get xy+yz+zx=−3 .

xy + yz + zx = -3

Start by cubing (x+y+z)

1 = x^3+y^3+z^3 + 3(x^2 y + y^2 x + x^2 z + z^2 x + y^2 z + z^2 y) + 6 xyz

1 = 13 + 3(xy(x+y) + xz(x+z) + yz(y+z)) + 6xyz

0 = 12 + 3(xy(x+y+z) + xz(x+y+z) + yz(x+y+z) -3xyz) + 6xyz

0 = 12 + 3(xy+xz+yz)(x+y+z) - 3xyz

0 = 12 +3(xy+xz+yz) - 3

xy+xz+yz = (3–12)/3 = -3

Please mark me as Brainlist and Follow.me

Answer 2

Answer:

-16

Step-by-step explanation:

hope this will help you