Question

if x+y+z=8 and xy+yz+zx =20 find the value of x ^3 +y^3 +z^3-3xyz

Answer 1

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
+x+y+z=8
+xy+yz+zx=20
8(x^2+y^2+z^2-20)
now firstfind the value of x^2+y^2+z^2
so for that put the identity(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
8^2=x^2+y^2+z^2+2(20)
24=x^2+y^2+z^2
now back to the part...
8(24-20)
=8×4
=32

Answer 2

$x + y + z =8 \\ squaring \: both \: the \: sides \\ (x + y + z) {}^{2} = (8) {}^{2} \\ x {}^{2} + y {}^{2} + z {}^{2} + 2(xy + yz + zx) = 64 \\ here \: xy + yz + xz = 20 \\ x {}^{2} + y {}^{2} + z {}^{2} + 2(20) = 64 \\ x {}^{2} + y {}^{2} + z {}^{2} + 40 = 64 \\ x {}^{2} + y {}^{2} + z {}^{2} = 64 - 40 \: ( = 24) \\ x {}^{2} + y {}^{2} + z {}^{2} = 24 \\ \\ \\ \\ x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = (x + y + z)(x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - xz) \\ (8)(24 - (20) \\ 8 \times 4 = 32$
mark me as brainlist please