Question

Solve for $(x,y)$:

$4x^3+3x^{2}y+y^3=8$,
$2x^3-2x^{2}y+x{y}^2=1$

• A. $(1,1)$
• B. $(\frac{1}{2},\frac{\sqrt{3}}{2})$
• C. $(2,\frac{3}{4})$
• D. $(5,\frac{2}{3})$

Answer 1

Answer: (A)

$(1,1)$

Put $y=mx$, and substitute in both equations.

Thus $x^3(4+3m+m^3)=8$ ..... (1)
$x^3(2-2m+m^2)=1$ .....(2)
Therefore, $\frac{4+3m+m^3}{2-2m+m^2}=8$;
$\Rightarrow m^3-8m^2+19m-12=0$
That is, $(m-1)(m-3)(m-4)=0$;
$\therefore m=1$, or $3$, or $4$.
(i) Take $m=1$, and substitute in either (1) or (2)
From (2), $x^3=1$

$\therefore x=1$
and $y=mx=x=1$.
(ii) Take $m=3$, and substitute in (2)
Thus $5x^3=1;\therefore x=\sqrt[3]{3\frac{1}{5}}$;
and $y=mx=3x=3\sqrt[2]{2\sqrt{1}5}$.
(iii) Take $m=4$; we obtain
$10x^3=1$

$\Rightarrow x=\sqrt[3]{3\frac{1}{10}}$;
and $y=mx=4x=4\sqrt[3]{3\frac{1}{10}}$.
Hence the complete solution is
$x=1,\sqrt[3]{3\frac{1}{5}},\sqrt[3]{3\frac{1}{10}}$.
$y=1,3\sqrt[3]{3\frac{1}{5}},4\sqrt[3]{3\frac{1}{10}}$.