$(x, y) \in$ $\mathbb{R} $, $x^2 +y^2 +xy=4$ and $x^4 +y^4 +(xy) ^2 =8$ calculate : $ x^6+(xy)^3+y^6$

Question

I did the answer, but I want new ideas for answer

My answer :

$x^2 +y^2 =4-xy $ $\Rightarrow $ $(*)$ $( x^2 +y^2) ^2 =(4-xy) ^2$

$ x^4 +y^4+2 (xy) ^2 = 8+(xy)^2$ $ \Rightarrow$ $(**)$ $(x^2 +y^2) ^2 =8+(xy) ^2$

$(*, **) $ $\Rightarrow$ $8+(xy) ^2 =(4-xy)^2$ $\Rightarrow$ $ 8+(xy) ^2= 16-8xy+(xy)^2$ $\Rightarrow$ (***) $xy=1$

$(***)$ $\Rightarrow$ $ x^2 +y^2 =3$ $\Rightarrow$ $ x^2 =3-y^2$  and $ y^2 = 3-x^2 $ $\Rightarrow$ $ x^6 =(3-y^2) ^3 and y^6 =(3-x^2) ^3 $ $\Rightarrow$ $ x^6 +y^6 =(3-y^2)^3 +(3-x^2) ^3$ $\Rightarrow$ $ x^6 +y^6 = - x^6 + 9 x^4 - 27 x^2 - y^6 + 9 y^4 - 27 y^2 + 54 $ $\Rightarrow $ $2(x^6 +y^6) = 9(x^4 +y^4) - 27(x^2 +y^2) +54 $ $\Rightarrow$ $ x^6 +y^6 = \frac{9(7)-27(3)+54}{2}= 18 $

Finally :

$x^6 +y^6 +(xy) ^3 =19 $

Because $xy =1$ and$ x^6+y^6 =18$

Answer 1

Hint

$$x^4+y^4+x^2y^2=(x^2+y^2)^2-(xy)^2=?$$

So, we have $x^2+y^2=?$ and $xy=?$

We don't need to find $x,y$

Now $$x^6+y^6+x^3y^3=(x^2+y^2)^3-3x^2y^2(x^2+y^2)+(xy)^3=?$$

Answer 2

$$8=x^4+x^2y^2+y^4=(x^2+y^2)^2-x^2y^2=$$ $$=(x^2-xy+y^2)(x^2+xy+y^2)=4(x^2+y^2-xy),$$ which gives $$x^2+y^2-xy=2,$$ $$x^2+y^2=3,$$$$xy=1$$ and $$x^6+y^6+x^3y^3=(x^2+y^2)(x^4-x^2y^2+y^4)+x^3y^3=3(8-2)+1=19.$$