#### Answer

$x\sqrt6$

#### Work Step by Step

$\frac{-3\sqrt{14x^3}}{-\sqrt{21x}}\longrightarrow$ factor the radicands; use mental math to see that 7x is a factor in both radicands
=$\frac{-3\sqrt{2\times{x^2}\times7x}}{-\sqrt{3\times{7x}}}\longrightarrow$ multiplication property of square roots
=$\frac{-3\times\sqrt2\times\sqrt{x^2}\times\sqrt{7x}}{-\sqrt3\times\sqrt{7x}}\longrightarrow$ simplify; $\sqrt{7x}$ cancels; negative cancels
=$\frac{-3\times\sqrt2\times x}{-\sqrt3}\longrightarrow$ simplify
=$\frac{3x\sqrt2}{\sqrt3}\longrightarrow$ multiply numerator and denominator by $\sqrt3$
=$\frac{3x\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\longrightarrow$ simplify; multiplication property of square roots
=$\frac{3x\sqrt6}{3}\longrightarrow$ simplify; 3 cancels
=$x\sqrt6$