+285 Given that
\((x+y+z)(xy+xz+yz)=18\)
and that
\(x^2(y+z)+y^2(x+z)+z^2(x+y)=6\)
for real numbers x, y, and z, what is the value of xyz?
-hihihi
😎😎😎
+285 (x+y+z)(xy+xz+yz)=25, x^2(y+z)+y^2(x+z)+z^2(x+y)=6
Expand the first equation
yx^2 + zx^2 + xyz + xy^2 + xyz + zy^2 + xyz + xz^2 + yz^2 = 18 simplify
yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2 + 3xyz = 6 (1)
Expand the second equation
yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2 = 6 (2)
Substitute (2) into (1)
6 + 3xyz = 18 subtract 6 from 18
3xyz = 12 divide both sides by 3
xyz = 4
-hihihi
😎😎😎
