Answer:
4
Step-by-step explanation:
[tex](x+y+z)(xy+xz+yz)=18[/tex] Equation 1
[tex]x^2(y+z)+y^2(x+z)+z^2(x+y)=6[/tex] Equation 2
What is the value of [tex]xyz[/tex] where each variable represents a real number?
Let's expand equation 1:
[tex](x+y+z)(xy+xz+yz)=18[/tex]
[tex]x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18[/tex]
Simplify each term if can:
[tex]x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18[/tex]
See if we can factor a little to get some of the left hand side of equation 2:
The first two terms have [tex]x^2[/tex] and if I factored [tex]x^2[/tex] from first two terms I would have [tex]x^2(y+z)[/tex] which is the first term of left hand side of equation 2.
So let's see what happens if we gather the terms together that have the same variable squared together.
[tex]x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18[/tex]
Factor the variable squared terms out of each binomial pairing:
[tex]x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18[/tex]
Replace the sum of those first three terms with what it equals which is 6 from the equation 2:
[tex]6+xyz+xyz+xyz=18[/tex]
Combine like terms:
[tex]6+3xyz=18[/tex]
Subtract 6 on both sides:
[tex]3xyz=12[/tex]
Divide both sides by 3:
[tex]xyz=4[/tex]
