Show/explain √6(3+2√2)²= and √3(√2-1)²=?

Question

simplify:
1) √6(3+2√2)²=

2) √3(√2-1)²=

Answer 1

1.
sqrt6(3 +2sqrt2)^2 =
sqrt6(3 +2sqrt2)(3 +2sqrt2)
sqrt6[9 + 6sqrt2 + 6sqrt2 + 4sqrt(2 x 2)]
sqrt6 [9 + 12sqrt2 + 4sqrt4]
sqrt6[9 +12sqrt2 + (4 x 2)]
sqrt6[9 +12sqrt2 +8]
sqrt6[17 + 12sqrt2]
17sqrt6 +12sqrt2sqrt6
17sqrt6 +12sqrt(2 x 2 x3)
17sqrt6 +24sqrt3
sqrt3[17sqrt2 + 24]

2.
sqrt3[sqrt2 - 1]^2
sqrt3[(sqrt2 -1)(sqrt2 - 1)]
sqrt3[sqrt2sqrt2 -sqrt2 - sqrt2 +1]
sqrt3[2 -2sqrt2 +1]
sqrt3[3 - 2sqrt2]
3sqrt3 - 2sqrt2sqrt3
3sqrt3 - 2sqrt6

Answer 2

Problem 1
sqrt(6)[(3 + 2sqrt(2)]^2 =
sqrt(6)[9 + (4)(2) + 2(3)(2)sqrt(2)]=
sqrt(6)[9 + 8 + 12 sqrt(2)] =
sqrt(6)[17 + 12 sqrt(2)] =
17 sqrt(6) + 12(sqrt(12) =
17 sqrt (6) + 12(sqrt(4)(3)) =
17 sqrt (6) + 24 sqrt(3) =
17 sqrt(3)(2) + 24 sqrt(3) =
sqrt (3)[24 + 17 sqrt(2)
Problem 2
sqrt(3)[sqrt(2) - 1]^2 =
sqrt(3)[2 + 1 - 2(1)sqrt(2)=
sqrt(3)[3 - 2 sqrt(2) =
3 sqrt(3) - 2 sqrt(6) =
3 sqrt(3) - 2 [sqrt(2)(3)]=
sqrt(3)[3 - 2 sqrt(2)]