Tìm x, y biết x^3 + y^3 = 9xy

Lời giải:

Ta có: \(x^3+y^3=9xy\)

\(\Leftrightarrow (x+y)^3-3xy(x+y)=9xy\)

\(\Leftrightarrow (x+y)^3-3xy(x+y+3)=0\)

\(\Leftrightarrow (x+y)^3+3^3-3xy(x+y+3)=27\)

\(\Leftrightarrow (x+y+3)[(x+y)^2-3(x+y)+9]-3xy(x+y+3)=27\)

\(\Leftrightarrow (x+y+3)[(x+y)^2-3(x+y)+9-3xy]=27\)

\(\Leftrightarrow (x+y+3)(x^2+y^2+9-xy-3x-3y)=27\)

Vì \(x,y\in\mathbb{N}^*\Rightarrow x+y+3\geq 5\)

Đến đây ta xét các TH:
TH1: \(\left\{\begin{matrix} x+y+3=9(1)\\ x^2+y^2+9-xy-3x-3y=3(2)\end{matrix}\right.\)

\((1)\rightarrow x+y=6\)

Thay vào PT thứ 2:

\((x+y)^2-2xy+9-xy-3(x+y)=3\)

\(\Leftrightarrow 27-3xy=3\Leftrightarrow xy=8\)

Thay \(y=6-x\Rightarrow x(6-x)=8\)

\(\Leftrightarrow x^2-6x+8=0\)

\(\Leftrightarrow (x-2)(x-4)=0\Leftrightarrow x=2, x=4\)

\(\Rightarrow y=4, y=2\)

TH2: \(\left\{\begin{matrix} x+y+3=27(1)\\ x^2+y^2+9-xy-3x-3y=1(2)\end{matrix}\right.\)

\((1)\rightarrow x+y=24\)

Thay vào (2):

\((x+y)^2-2xy+9-xy-3(x+y)=1\)

\(\Leftrightarrow 513-3xy=1\Leftrightarrow xy=\frac{512}{3}\not\in\mathbb{N}^*\) (loại)

Vậy \((x,y)=(2,4); (4,2)\)