1. **State the problem:** Express $8 - 3\sqrt{6}$ in the form $m\sqrt{3} + n\sqrt{2}$ where $m$ and $n$ are rational numbers. 2. **Recall the form:** We want to find rational numbers $m$ and $n$ such that: $$8 - 3\sqrt{6} = m\sqrt{3} + n\sqrt{2}$$ 3. **Rewrite $\sqrt{6}$:** Note that $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$. 4. **Substitute:** The equation becomes: $$8 - 3\sqrt{2}\sqrt{3} = m\sqrt{3} + n\sqrt{2}$$ 5. **Group terms:** To express the left side in terms of $\sqrt{3}$ and $\sqrt{2}$, write: $$8 = 0\sqrt{3} + 0\sqrt{2} + 8$$ But since the right side has no constant term, the constant 8 must be represented as a combination of $\sqrt{3}$ and $\sqrt{2}$, which is impossible unless $m$ and $n$ include irrational parts. So the problem likely means to express $8 - 3\sqrt{6}$ as $m\sqrt{3} + n\sqrt{2}$ plus a rational constant. 6. **Assuming the problem means:** $$8 - 3\sqrt{6} = p + m\sqrt{3} + n\sqrt{2}$$ where $p, m, n$ are rational numbers. 7. **Set up the equation:** $$8 - 3\sqrt{2}\sqrt{3} = p + m\sqrt{3} + n\sqrt{2}$$ 8. **Equate rational and irrational parts:** - Rational part: $8 = p$ - Coefficient of $\sqrt{3}$: $-3\sqrt{2} = m\sqrt{3}$ ??? This is not directly comparable. 9. **Alternative approach:** Let’s try expressing $8 - 3\sqrt{6}$ as $(a + b\sqrt{2})(c + d\sqrt{3})$ where $a,b,c,d$ are rational. 10. **Expand:** $$(a + b\sqrt{2})(c + d\sqrt{3}) = ac + ad\sqrt{3} + bc\sqrt{2} + bd\sqrt{6}$$ 11. **Match terms:** We want: $$ac = 8$$ $$ad = m$$ (coefficient of $\sqrt{3}$) $$bc = n$$ (coefficient of $\sqrt{2}$) $$bd = -3$$ (coefficient of $\sqrt{6}$) 12. **Choose $a=2$, $c=4$ to satisfy $ac=8$.** 13. **From $bd = -3$, pick $b= -1$, $d=3$ (both rational).** 14. **Calculate $ad = 2 \times 3 = 6$ and $bc = -1 \times 4 = -4$.** 15. **Therefore:** $$8 - 3\sqrt{6} = (2 - \sqrt{2})(4 + 3\sqrt{3})$$ 16. **Expand to verify:** $$2 \times 4 = 8$$ $$2 \times 3\sqrt{3} = 6\sqrt{3}$$ $$-\sqrt{2} \times 4 = -4\sqrt{2}$$ $$-\sqrt{2} \times 3\sqrt{3} = -3\sqrt{6}$$ Sum: $$8 + 6\sqrt{3} - 4\sqrt{2} - 3\sqrt{6}$$ 17. **But original expression is $8 - 3\sqrt{6}$, so to get rid of $6\sqrt{3}$ and $-4\sqrt{2}$, we need to adjust.** 18. **Hence, the expression cannot be written exactly as $m\sqrt{3} + n\sqrt{2}$ without a constant term.** 19. **Final conclusion:** $$8 - 3\sqrt{6} = 8 + 0\sqrt{3} + 0\sqrt{2} - 3\sqrt{6}$$ It cannot be expressed purely as $m\sqrt{3} + n\sqrt{2}$ with rational $m,n$ without including the constant 8 and the $\sqrt{6}$ term. **If the problem intends to express $\sqrt{6}$ in terms of $\sqrt{3}$ and $\sqrt{2}$, then:** $$\sqrt{6} = \sqrt{2} \times \sqrt{3}$$ So the expression is already in simplest form. **Therefore, the answer is:** $$8 - 3\sqrt{6} = 8 - 3\sqrt{2}\sqrt{3}$$ No simpler form $m\sqrt{3} + n\sqrt{2}$ with rational $m,n$ exists. **Slug:** express root **Subject:** algebra