1. Stating the problem: Simplify the expression $$\frac{\sqrt{-6}}{\sqrt{-3} \times \sqrt{-5}}$$ and write the result in the form $a + bi$ where $a$ and $b$ are real numbers. 2. Recall that for any negative number under a square root, we use the imaginary unit $i$ where $i = \sqrt{-1}$. So, $\sqrt{-x} = \sqrt{x}i$ for $x > 0$. 3. Rewrite each square root: $$\sqrt{-6} = \sqrt{6}i$$ $$\sqrt{-3} = \sqrt{3}i$$ $$\sqrt{-5} = \sqrt{5}i$$ 4. Substitute these into the expression: $$\frac{\sqrt{6}i}{\sqrt{3}i \times \sqrt{5}i}$$ 5. Simplify the denominator: $$\sqrt{3}i \times \sqrt{5}i = \sqrt{3} \times \sqrt{5} \times i \times i = \sqrt{15} \times i^2$$ 6. Since $i^2 = -1$, the denominator becomes: $$\sqrt{15} \times (-1) = -\sqrt{15}$$ 7. Now the expression is: $$\frac{\sqrt{6}i}{-\sqrt{15}} = -\frac{\sqrt{6}i}{\sqrt{15}}$$ 8. Simplify the fraction by rationalizing the denominator: $$-\frac{\sqrt{6}i}{\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}} = -\frac{\sqrt{6} \times \sqrt{15} i}{15}$$ 9. Multiply the square roots in the numerator: $$\sqrt{6} \times \sqrt{15} = \sqrt{6 \times 15} = \sqrt{90}$$ 10. Simplify $\sqrt{90}$: $$\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}$$ 11. Substitute back: $$-\frac{3\sqrt{10} i}{15}$$ 12. Simplify the fraction: $$-\frac{3\sqrt{10} i}{15} = -\frac{\cancel{3}\sqrt{10} i}{\cancel{15}5} = -\frac{\sqrt{10}}{5} i$$ 13. Write in the form $a + bi$: $$0 - \frac{\sqrt{10}}{5} i$$ 14. Therefore, the real part $a = 0$ and the imaginary part $b = -\frac{\sqrt{10}}{5}$. Final answer: $$0 - \frac{\sqrt{10}}{5} i$$