高中一年級數學問題，請問怎麼解答?謝謝?

Sol
a+b>0
a－b>0
a^2+2ab+b^2=3√3－√2
a^2－2ab+b^2=3√2－√3
4ab=4√3－4√2
ab=√3－√2
a^2b^2=5－2√6
a^2－b^2=√[(3√3－√2)*(3√2－√3)]
=√(9√6－9－6+√3)
=√(10√6－15)
a^4+a^2b^2+b^4
=(a^4－2a^2b^2+b^4)+3a^2b^2
=10√6－15+3*(5－2√6)
=4√6

a+b=√(3√3-√2)，a=√(3√3-√2)-b
帶入a-b=√(3√2-√3)
√(3√3-√2)-b-b=√(3√2-√3)
-2b=√(3√2-√3)-√(3√3-√2)
b=[√(3√3-√2)-√(3√2-√3)]/2

a+b=√(3√3-√2)，b=√(3√3-√2)-a
帶入a-b=√(3√2-√3)
a-[√(3√3-√2)-a]=√(3√2-√3)
2a=√(3√3-√2)+√(3√2-√3)
a=[√(3√3-√2)+√(3√2-√3)]/2

ab=[√(3√3-√2)+√(3√2-√3)]/2*[√(3√3-√2)-√(3√2-√3)]/2
=[3√3-√2-(3√2-√3)]/4=(4√3-4√2)/4=√3-√2

a^2+b^2=(a+b)^2+(a-b)^2=(3√3-√2)+(3√2-√3)=2√3+2√2

a^4+(ab)^2+b^4=a^4+2(ab)^2+b^4-(ab)^2=(a^2+b^2)^2-(ab)^2=(a^2+b^2-ab)(a^2+b^2+ab)
=(2√3+2√2+√3-√2)(2√3+2√2-√3+√2)=(3√3+√2)(√3+3√2)=9+9√6+√6+6=15+10√6

(a+b)^2 = 3√3 - √2 ..... (1)
(a-b)^2 = 3√2 - √3 ..... (2)

(1) - (2) 得:
4ab = 2√3 + 2√2
ab = (1/2)( √3 + √2 )
a^2b^2 = (1/4)( 5 + 2√6 ) ..... (3)

(1)*(2) 得:
( a^2 - b^2 )^2 = 9√6 - 9 - 6 + √6 = 10√6 - 15
a^4 - 2a^2b^2 + b^4 = 10√6 - 15 ..... (4)

a^4 + a^2b^2 + b^4
= ( a^4 - 2a^2b^2 + b^4 ) + 3a^2b^2
= 10√6 - 15 + 3*(1/4)( 5 + 2√6 ) , 由 (3), (4) 得
= 10√6 - 15 + 15/4 + (3/2)√6
= (23/2)√6 - 45/4 ..... Ans