The previous tutor found the solution by the method of "trial and error". The question remains still open if it can be solved on the more solid algebraic base. The answer is "Yes", and I will show you "How". With one apology: since the problem is slightly higher than the traditional school math, the solution is, correspondingly, slightly higher (but still understandable).
We are given the numerical values for the functions, and . A remarkable fact is that the function can be explicitly expressed via , and . (I will do it later). So, knowing the numerical values of , and , we can find the value of . Next, knowing the numerical values of , and , we can consider the polynomial of the variable "u" P(u) = . (2) This polynomial is nothing else as P(u) = (u-x)*(u-y)*(u-z). (3) In other words, the polynomial (2) is factorable into (3). Now, instead of solving the system (1), we can solve the polynomial equation P(u) = 0 (4) for u. If we solve it (and when we solve it), its roots u = x, u = y and u = z will be the solution of the system (1). Why this way is better than solving (1) directly? Well, for example, you can (try) to solve the polynomial equation (4) graphically. Or apply other methods specific for polynomial equations. You will see it later. Now I will implement this methodology.
We will do it step by step: 1.= 2. = = x(xy+xz) + y(xy+yz) + z(xz+yz) = = x(xy+xz+yz-yz) + y(xy+yz+xz-xz) + z(xz+yz+xy-xy) = x(xy+xz+yz) + y(xy+xz+yz) + z(xy+xz+yz) - x(yz) - y(xz) - z(xy) = (x+y+z)(xy+xz+yz) - 3xyz 3. Therefore, = , or = , or = , or xyz = . 4. Thus xyz = = = 48.
So, our polynomial (2) is P(u) =, and we need solve this polynomial equation (4) = . First, let's do it graphically.
Figure. Plot P(u) = |
Do you see the roots? Of course, they are u=2, u=4 and u=6. And you can check it manually substituting these values into the polynomial. Or you can apply the rational roots theorem. All the roots are among the integer divisors of the number 48, and you have only finite number of options to check. It is your other method to find the roots.