Can we compare them?

$\begin{aligned} \frac {\sqrt{12}-\sqrt{27}} {6} \times \sqrt{3}\times \sqrt{4} & = \frac {(2\sqrt{3}-3\sqrt{3})\sqrt{3}\cdot{}2} {6} \\ & = \frac {(-\sqrt{3})\sqrt{3}} {3} \\ & = \frac {-3}{3} \\ & = \boxed{-1} \end{aligned}$

$\frac{\sqrt{12}-\sqrt{27}}{6}\times\sqrt{3}\times\sqrt{4}$

$= \frac{\sqrt{3}(\sqrt{4}-\sqrt{9})}{6}\times\sqrt{3}\times\sqrt{4}$

$= \frac{\sqrt{4}-\sqrt{9}}{6}\times3\times2$

$= \frac{\sqrt{4}-\sqrt{9}}{6}\times6$

$= \sqrt{4}-\sqrt{9} = 2-3 = \boxed{-1}$

$\dfrac{\sqrt{12}-\sqrt{27}}{6}\times\sqrt{3}\times\sqrt{4}$

First we turn them all into radical form,

$\left(\sqrt{\dfrac{12}{36}}-\sqrt{\dfrac{27}{36}}\right)\times\sqrt{12}$

$=\sqrt{\frac{144}{36}}-\sqrt{\frac{324}{36}}$

$=\sqrt{4}-\sqrt{9}$

$=2-3$

$=\boxed{-1}$

$\frac{\sqrt12-\sqrt27}{6}\times\sqrt3\times\sqrt4$
$=\frac{2\sqrt3-3\sqrt3}{\sqrt36}\times\sqrt3\times\sqrt4$
$=\frac{-\sqrt3}{2\times\sqrt3\times\sqrt3}\times\sqrt3\times2$
$=\boxed{-1}$

$\frac { \sqrt { 3 } \cdot \sqrt { 4 } -\sqrt { 3 } \cdot \sqrt { 9 } }{ 6 } \quad \times \quad \sqrt { 3 } \quad \times \sqrt { 4 }$

$=\frac { \sqrt { 3 } (\sqrt { 3 } \cdot 2-\sqrt { 3 } \cdot 3) }{ 6 } \quad \quad \times \quad 2$

$=\frac { ((3\cdot 2)-(3\cdot 3)) }{ 6 } \quad \quad \times \quad 2$

$=\frac { 6-9 }{ 3 } \quad =\quad -1$

hello see this java code:

public class IAmBrilliant {
 public static void main(String[] args){
  System.out.println((Math.sqrt(12)-Math.sqrt(27))/6*Math.sqrt(3)*Math.sqrt(4));
 }
}

it outputs:

-1.0

anyone know how to do it using math?

$\qquad \frac { \sqrt { 12 } -\sqrt { 27 } }{ 6 } \times \sqrt { 3 } \times \sqrt { 4 } \\ =\quad \quad \frac { 2\sqrt { 3 } -3\sqrt { 3 } }{ 6 } \times 2\sqrt { 3 } \\ =\quad \quad \frac { (2-3)\sqrt { 3 } }{ 6 } \times 2\sqrt { 3 } \\ =\quad \quad \frac { -\sqrt { 3 } }{ 6 } \times 2\sqrt { 3 } \\ =\quad \quad \frac { -2(3) }{ 6 } \qquad \qquad \because \sqrt { 3 } \times \sqrt { 3 } =3\\ =\quad \quad \frac { -6 }{ 6 } \quad =\quad \boxed { -1 }$

$1. \frac{\sqrt{12} - \sqrt{27}}{6} \times \sqrt{3} \times \sqrt{4} - Given$ $2. =\frac{\sqrt{4 \times 3} - \sqrt{9 \times 3}}{6}\times\sqrt{3} \times 2 -Break Radicals Into Squares$ $3. = \frac{\sqrt{4}\times\sqrt{3} - \sqrt{9}\times\sqrt{3}}{6}\times2\sqrt{3} - Simplify Radicals$
$4. =\frac{2\sqrt{3} - 3\sqrt{3}}{6}\times 2\sqrt{3} - Simply Radicals$ $5. =\frac{-1\sqrt{3}}{6}\times\frac{2\sqrt{3}}{1} - Combine Like Terms$ $6. =\frac{-1\sqrt{3}\times2 \sqrt{3}}{6} - Multiply Fractions$ $7. =\frac{-1\times2\times\sqrt{3}\times \sqrt{3}}{6} - Commutative Property$
$8. =\frac{-2\times 3}{6} = \frac{-6}{6} = \boxed{-1} -Simplify$

$\frac {\sqrt{12}-\sqrt{27}}{6} \times \sqrt{3} \times \sqrt{4}$

= $\frac {\sqrt{3 \times 4}-\sqrt{3 \times 9}}{2 \times 3} \times \sqrt{3} \times 2$

= $\frac {2\sqrt{3}-3\sqrt{3}}{3} \times \sqrt{3}$

= $-\frac {\sqrt{3}}{3} \times \sqrt{3}$

= $-\frac {3}{3}=\boxed {-1}$

It's simple guys. Just convert all roots to whole numbers(i.e √4 = 2) and then convert 6 to corresponding roots. Then you'll just need to use some sense to simplify the answer.( I am a 10th grader and did this sum in 20 secs)

$\sqrt{12} = 2 \sqrt{3}$

$\sqrt{27} = 3 \sqrt{3}$

$2\sqrt3 - 3\sqrt3 = -1\sqrt{3}$

$\sqrt{3} \times \sqrt{4} = \sqrt{12} = 2\sqrt{3}$

$\frac{ 2 \sqrt{3} } {6} = \frac{\sqrt{3}}{3}$

$\frac{\sqrt{3}}{3} \times -1\sqrt{3} = \frac{-3}{3} = -1$

There is no need of solution its a basic concept which only need a little focus

((√12 - √27) / 6 ) x √3 x √4 = ( (2√3-3√3) x √3x2 ) / 6 = ( 4√9 - 6√9 ) / 6 = (4x3 - 6x3)/ 6 = ( 12 - 18 ) / 6 = -6 /6 = -1

Its amazing how many different ways there are to solve this problem. And this is the reason why math is such a beautiful subject.

$\frac{\sqrt{12}-\sqrt{27}}{6} \times \sqrt{3} \times \sqrt{4}=\sqrt{12} \times \frac{\sqrt{12}(2\sqrt{3}-3\sqrt{3})}{6}=\frac{\sqrt{12} \times -\sqrt{3}}{6}=\frac{-\sqrt{36}}{6}=\frac{-6}{6}=\boxed{\large{-1}}$