Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?

Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?

$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$ but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.

Also, when $a=b\ne 0$, $(a^2+ab+b^2)(a-b) = a^3-b^3 =0$.

Hint :

$$a^2 + ab + b^2 = \left( a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}$$

Fix $b\neq0$ and try to solve in $a$. This a second order equations so the possible values of $a$ in terms of $b$ are

$$ a_{1,2}=\frac{-b\mp \sqrt{-3b^2}}{2}$$

So, for any non zero real value $b$ the possible values of $a$ are complex conjugated because $-3b^2<0$. Therefore you cannot find two non-zero reals such that $(a+b)^2=ab$.