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yaroslaw [1]
3 years ago
8

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?

Mathematics
1 answer:
worty [1.4K]3 years ago
4 0

Answer:

4

Step-by-step explanation:

(x+y+z)(xy+xz+yz)=18 Equation 1

x^2(y+z)+y^2(x+z)+z^2(x+y)=6 Equation 2

What is the value of xyz where each variable represents a real number?

Let's expand equation 1:

(x+y+z)(xy+xz+yz)=18

x(xy)+x(xz)+x(yz)+y(xy)+y(xz)+y(yz)+z(xy)+z(xz)+z(yz)=18

Simplify each term if can:

x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y=18

See if we can factor a little to get some of the left hand side of equation 2:

The first two terms have x^2 and if I factored x^2 from first two terms I would have x^2(y+z) 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.

x^2y+x^2z+y^2x+y^2z+z^2y+z^2x+xyz+xyz+xyz=18

Factor the variable squared terms out of each binomial pairing:

x^2(y+z)+y^2(x+z)+z^2(y+x)+xyz+xyz+xyz=18

Replace the sum of those first three terms with what it equals which is 6 from the equation 2:

6+xyz+xyz+xyz=18

Combine like terms:

6+3xyz=18

Subtract 6 on both sides:

3xyz=12

Divide both sides by 3:

xyz=4

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