Ta có
$2.x^3-1 = 15$
$<-> 2.x^3 = 16$
$<-> x^3 = 8$
$<-> x^3 = 2^3$
$<-> x = 2$.
Theo đề bài ta có
$x + \dfrac{16}{9} = y - \dfrac{25}{16}$
$<-> 2 + \dfrac{16}{9} + \dfrac{25}{16} = y$
$<-> y = \dfrac{769}{144}$
Lại có
$x + \dfrac{16}{9} = z + \dfrac{9}{25}$
$<-> 2 + \dfrac{16}{9} - \dfrac{9}{25} = z$
$<-> z = \dfrac{769}{225}$
Vậy $(x, y, z) = \left( 2, \dfrac{769}{144}, \dfrac{769}{225} \right)$.
