3. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$

Then the range of values for $t$ is $\qquad$
Hint: The answer is $-3 \leqslant t \leqslant-\frac{1}{3}$.