依题意
[6-2b√(a+3b)]^2=a+3b,①
[2a+(1-a^2)^(1/3)]^3=1-a^2,②(改题了)
由②,8a^3+12a^2*(1-a^2)^(1/3)+6a(1-a^2)^(2/3)+(1-a^2)=1-a^2,
∴a=0,或4a^2+6a(1-a^2)^(1/3)+3(1-a^2)^(2/3)=0(无实根)
代入①,[6-2b√(3b)]^2=3b,
36-24b√(3b)+12b^3=3b,
4b^3-b+12=8b√(3b),③
平方得16b^6-8b^4+96b^3+b^2-24b+144=192b^3,
∴16b^6-8b^4-96b^3+b^2-24b+144=0,
解得b1≈1.12963765,b2≈1.79337634.
题目似乎有些难!