Question

X+y+z=6, xy+yz+zx=10, find x^3+ y^3+z^3=3xyz

Answer 1

Answer:

$x^{3}+y^{3}+z^{3}-3xyz=(x+y+z) (x^{2}+y^{2}+z^{2}+xy+yz+xz) \\ = 6 \times (x^{2}+y^{2}+z^{2} + 10)$

we need to find x^2+y^2+z^2

$(x + y + z)^{2} = x^{2}+y^{2} + z^{2} + 2(xy +zy+xz) \\ {6}^{2} = x^{2}+y^{2}+z^{2} + 2 \times 10 \\ 36 =x^{2}+y^{2}+z^{2}+2 \times10$

$x^{2}+y^{2}+z^{2} = 36 - 20 \\ x^{2}+y^{2}+z^{2} = 16$

now put the value

$x^{3}+y^{3}+z^{3}-3xyz = (x+y+z) (x^{2}+y^{2}+z^{2}+xy+yz+xz) \\ x^{3}+y^{3}+z^{3}-3xyz = 6 \times (16 + 10) \\ x^{3}+y^{3}+z^{3}-3xyz =6 \times 160= 960$

hope you like the answer...