what is answer??????????????????

$$\\\sqrt3(1+\sqrt{2})*\sqrt{6}(3-2*\sqrt{2})\\\\ =\sqrt{3*6}*[(1+\sqrt{2})(3-2\sqrt{2})]\\\\ =\sqrt{2*9}*[1(3-2\sqrt{2})+\sqrt{2}(3-2\sqrt{2})]\\\\ =3\sqrt{2}*[3-2\sqrt{2}+3\sqrt{2}-2\sqrt{2}\sqrt{2}]\\\\ =3\sqrt{2}*[3+\sqrt{2}-2*2]\\\\ =3\sqrt{2}*[\sqrt{2}-1]\\\\ =6-3\sqrt{2}$$

You need to check this answer :)

I assume this is:

sqrt3(1+sqrt(2))*sqrt6(3-2*sqrt(2))   ....so we have

√3(1 + √2 ) ( √6)(3 - 2√2)  =

√18( 1 + √2 ) ( 3 - 2√2 ) =

√18 ( 3 + 3√2 - 2√2 - 4) =

√18( √2 - 1) =

√36  - √18 =

6 - 3√2

It's correct, Melody.....

Thanks Chris :)