Question

If x+y+z = 8 and xy + yz + zx = 20, find the value of x3 + y3 + z3 - 3xyz

Answer 1

in this case 2 algebric Identity are used
${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2( xy + yz + xz) \\ put \: the \: values \\ {8}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(20) \\ 64 = {x}^{2} + {y}^{2} + {z}^{2} + 40 \\ {x}^{2} + {y}^{2} + {z}^{2} = 24 \\ now \: another \: identity \\ {x}^{2} + {y}^{2} + {z}^{2} - 3xyz = (a + b + c)( {x}^{2} + {y}^{2} + {z}^{2} - ( xy+ yz+zx ) ) \\ put \: values\\ {x}^{2} + {y}^{2} + {z}^{2} - 3xyz = (8)(24 - 20) \\ {x}^{2} + {y}^{2} + {z}^{2} - 3xyz = 8 \times 4 \\ {x}^{2} + {y}^{2} + {z}^{2} - 3xyz = 32 \\ mark \: as \: brainliest \: if \: you \: like \: it$

Answer 2

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
=>x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)
=>x^2+y^2+z^2=(8)^2-2(20)
=>x^2+y^2+z^2=64-40
=>x^2+y^2+z^2=24

Now,
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=(x+y+z){(x^2+y^2+z^2)-(xy+yz+zx)}
=8×(24-20)
=8×4
=32


Hence,x^3+y^3+z^3-3xyx=32