help hard system

Assuming $x$, $y$, and $z$ are positive real numbers satisfying
$xy - z = 15\\ xz - y = 15\\ yz - x = 15$
then, what is the value of $xyz$?

$\begin{array}{|lrcll|} \hline (1) & xy - z &=& 15 \\ (2) & xz - y &=& 15 \\ (3) & yz - x &=& 15 \\ \hline (1)+(2)+(3):& (xy - z)+(xz - y)+(yz - x) &=& 45 \\ & \ldots \\ &\mathbf{ x+y+z } &=& \mathbf{ xy+xz+yz-45 } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline yz - x &=& 15 \\ \mathbf{x} &=& yz-15 \\ \hline xz-y &=& 15 \\ (yz-15)z-y &=& 15 \\ yz^2-15z-y &=& 15 \\ yz^2-y &=& 15z+15 \\ y(z^2-1) &=& 15(z+1) \\ y(z-1)(z+1) &=& 15(z+1) \quad | \quad : (z+1),~ z>0!\\ y(z-1)&=& 15\\ \mathbf{y} &=& \mathbf{\dfrac{15} {z-1} }\qquad (4) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline xz - y &=& 15 \\ \mathbf{y} &=& xz-15 \\ \hline xy-z &=& 15 \\ (xz-15)x-z &=& 15 \\ zx^2-15x-z &=& 15 \\ zx^2-z &=& 15x+15 \\ z(x^2-1) &=& 15(x+1) \\ z(x-1)(x+1) &=& 15(x+1) \quad | \quad : (x+1),~ x>0!\\ z(x-1)&=& 15\\ \mathbf{z} &=& \mathbf{\dfrac{15} {x-1} }\qquad (5) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline xy - z &=& 15 \\ \mathbf{z} &=& xy-15 \\ \hline yz-x &=& 15 \\ (xy-15)y-x &=& 15 \\ xy^2-15y-x &=& 15 \\ xy^2-x &=& 15y+15 \\ x(y^2-1) &=& 15(y+1) \\ x(y-1)(y+1) &=& 15(y+1) \quad | \quad : (y+1),~ y>0!\\ x(y-1)&=& 15\\ \mathbf{x} &=& \mathbf{\dfrac{15} {y-1} }\qquad (6) \\ \hline \end{array}$

$\small{ \begin{array}{|lrcll|} \hline (4)\times(5)\times(6):& xyz &=& \dfrac{15} {y-1} \times \dfrac{15} {z-1} \times \dfrac{15} {x-1} \\\\ & xyz &=& \dfrac{15^3} {(x-1)(y-1)(z-1)} \\\\ &&& \boxed{\mathbf{(x-1)(y-1)(z-1)}\\=xyz-(xy+xz+yz)+(x+y+z) - 1} \\\\ & xyz &=& \dfrac{15^3} {xyz-(xy+xz+yz)+(x+y+z) - 1} \\\\ &&& \mathbf{ x+y+z=xy+xz+yz-45 } \\\\ & xyz &=& \dfrac{15^3} {xyz-(xy+xz+yz)+(xy+xz+yz-45) - 1} \\\\ & xyz &=& \dfrac{15^3} {xyz-46} \\\\ & xyz(xyz-46) &=& 15^3 \\\\ & (xyz)^2-46xyz- 15^3 &=& 0 \\ \hline & xyz &=& \dfrac{46\pm\sqrt{46^2-4*(-15^3)}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{46^2+4*15^3}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{15616}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{16^2*61}}{2} \\\\ & xyz &=& \dfrac{46\pm 16\sqrt{61}}{2} \\\\ & \mathbf{ xyz } &=& \mathbf{ 23 +8\sqrt{61} } \qquad xyz >0! \\ \hline \end{array} }$

xy  - z    =  15

xz  - y    =  15

yz  - x =     15

Subtract  the  2nd equation from the 1st

xy - xz  + y - z   =   0

x (y -z)  + 1(y -z)   =  0

(x + 1)  ( y - z)  =  0                          since x is positive, divide both sides  by  (x + 1)

y - z   =   0

y  = z

Subtract  the  3rd  equation from the  2nd

xz - yz  + x - y =  0

z (x - y)  + 1(x - y)   = 0

(z + 1) ( x - y)   = 0                  again, z is positive,  divide both sides by (z + 1)

x - y   = 0

x = y

So   x = y =  z

In the 1st equation  we  have  that

x^2  - x =  15

x^2 - x  + 1/4   =  15  + 1/4

(x - 1/2)^2  = 61/4

x - 1/2  =   sqrt (61) / 2

x =  ( 1 + sqrt (61) ) / 2

So

xyz =    [  ( 1 + sqrt (61) ) /2  ] ^3    =    23  +  8sqrt (61)