system

Hi Guest!

Rewrite the first two equations as follows:

\(xy=15+z\)  (1)

\(xz=15+y \)  (2)

Subtracting (2) from (1) gives:

\(x(y-z)=-(y-z)\)       (Notice: \(z-y=-(y-z)\) ).

\(x(y-z)+(y-z)=0\)        (Add (y-z) to both sides to make R.H.S = 0 , and factor (y-z)):

\((y-z)(x+1)=0\)

Thus, \(y=z, \text{or } x=-1 (rejected)\)  (As given x,y,z > 0).

Therefore, \(y=z\) only.

Also, using the second and third equation:

\(xz=15+y\)  (2)

\(yz=15+x\)  (3)

Subtracting (3) from (2) to get: 

\(z(x-y)=-(x-y)\)

\(z(x-y)+(x-y)=0\)

\((x-y)(z+1)=0\)

Therefore, \(x=y\)        (z>0 so z=-1 is rejected).

But we found that:  \(y=z\)

Thus, It must mean that they are all equal to each other. That is,  \(x=y=z\)

So our system of equations becomes:

\(x^2-x=15\) \(\iff x^2-x-15=0 \iff x=\frac{1+\sqrt{61}}{2}\)  (A quadratic equation in x, which can be solved by the quadratic formula.)

So the desired answer is:

\(xyz=x^3=\frac{(1+\sqrt{61})^3}{8}=23+8\sqrt{61}\)

I hope this helps!