Find the value of $x+y$ knowing that $x^3+y^3=9xy-27$

Question

Find the value of $x+y$

where $x^3+y^3=9xy-27.$

I tried using $a^3+b^3=(a+b)(a^2+b^2−ab)$ and also $a^3+b^3=(a+b)^3−3ab(a+b)$

but couldn't find the answer. Please explain how do I solve these types of questions.

Answer given = 6

Answer 1

Let's just consider real situation. Let $u=x+y$, $v=xy$, then $u^3-3uv=9v-27$, i.e., $(u+3)(u^2-3u+9-3v)=0$. So one case is $u=-3$, another is $u\neq -3$ and $v=\frac{1}{3}(u^2-3u+9)$. Since $x$, $y$ are both real, $u^2\ge 4v$, i.e., $(u-6)^2\le 0$, then $u=6$.

Answer 2

Hint: Use Euler's identity $$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\big( (x-y)^2+(x-z)^2+(y-z)^2\big)$$

Answer 3

If $x=y=3$ we have $x+y=6$.

Let $x\neq y$. Thus, $$x^3+y^3+27-9xy=(x+y+3)(x^2+y^2+9-xy-3x-3y),$$ which gives $x+y=-3$ because $x^2+y^2+9-xy-3x-3y\neq0$ for $x\neq y$.

Id est, the answer is $\{6,-3\}$.

Answer 4

Let

$$\tag{1} f(x,y):=x^3+y^3-9xy+27=0$$

I don't agree with the unique answer $x+y=6$ to this problem. It should also include $x+y=-3$.

Here is why:

We are looking for constant(s) $k$ such that:

$$\tag{2}x+y=k.$$

Let us substitute $y=k-x$ in (1), yielding the equivalent problem : find $k$ and $x$ such that:

$$\tag{3} 3(k+3) x^2 - 3k(k+3)x + (27+k^3)=0$$

But $27+k^3=3^3+k^3=(k+3)(k^2-3k+9)$; thus (3) is equivalent to;

$$\tag{4} (k+3)(3 x^2 - 3kx + (k^2-3k+9))=0$$

Two consequences:

• either $k=-3$ which is one possibility for $k$ ; and any x is solution, thus any $(x,y)=(-x,-3-x)$ is a solution.

• or $x$ and $k$ are such that $x$ is a real root of quadratic equation

$$\tag{5} 3 x^2 - 3kx + (k^2-3k+9)=0$$

whose discriminant: $\Delta=-3(k-6)^2$ is always negative but for the particular value $k=6$, in which case $x$ is equal to the double root of this equation, i.e. $x=\dfrac{1}{6}(3 \times 6+0)=3$; therefore, $y=k-x=6-3=3$.

Let us conclude:

• either $x+y=-3$, and there is an infinity of solutions $\forall x, (x,y)=(x,-3-x)$.

• or $x+y=6$, and there is a unique solution $(x_0,y_0)=(3,3).$

This is made very clear when we consider the "curve" of implicit function defined by (1), which is in two parts : a straight line with equation $x+y=-3$ representing the "general case" and an isolated point correponding to the particular solution $(x_0,y_0)=(3,3).$

Remark: Calling figure 1 a "curve" may look exaggerated. We must understand that it is a limit case of curves with equation $f(x,y)=z$ where $z$ is a real parameter; said otherwise of different level curves of the surface with equation $z=f(x,y)$ (see figure 2 for a representation of some of them). All are cubic curves. OIn the case $z=27$ (in red), thus with equation $x^3+y^3-9xy=0$, we have the folium of Descartes (https://en.wikipedia.org/wiki/Folium_of_Descartes).