a: Đặt \(B=\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}\)

\(=\left(\sqrt{1+x}+\sqrt{1-x}\right)\left\lbrack\left(\sqrt{1+x}\right)^2-\sqrt{1+x}\cdot\sqrt{1-x}+\left(\sqrt{1-x}\right)^2\right\rbrack\)

\(=\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(1+x-\sqrt{1-x^2}+1-x\right)=\left(2-\sqrt{1-x^2}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)\)

Ta có: \(A=\sqrt{1-\sqrt{1-x^2}}\cdot\frac{\left\lbrack\left(\sqrt{1+x}\right)^3+\sqrt{\left(1-x\right)^3}\right\rbrack}{2-\sqrt{1-x^2}}\)

\(=\sqrt{1-\sqrt{1-x^2}}\cdot\frac{\left(2-\sqrt{1-x^2}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{2-\sqrt{1-x^2}}\)

\(=\sqrt{1-\sqrt{1-x^2}}\cdot\left(\sqrt{1+x}+\sqrt{1-x}\right)\)

Đặt \(C=\sqrt{1+x}+\sqrt{1-x}\)

=>\(C^2=1+x+1-x+2\cdot\sqrt{\left(1+x\right)\left(1-x\right)}=2+2\cdot\sqrt{1-x^2}\)

=>\(C=\sqrt{2+2\cdot\sqrt{1-x^2}}\)

Ta có: \(A=\sqrt{1-\sqrt{1-x^2}}\cdot\left(\sqrt{1+x}+\sqrt{1-x}\right)\)

\(=\sqrt{1-\sqrt{1-x^2}}\cdot\sqrt2\cdot\sqrt{1+\sqrt{1-x^2}}=\sqrt2\cdot\sqrt{1^2-\left(1-x^2\right)}=\sqrt2\cdot\left|x\right|\)

b: \(a^3-a^2b+ab^2-6b^3=0\)

=>\(a^3-2a^2b+a^2b-2ab^2+3ab^2-6b^3=0\)

=>\(a^2\left(a-2b\right)+ab\left(a-2b\right)+3b^2\left(a-2b\right)=0\)

=>\(\left(a-2b\right)\left(a^2+ab+3b^2\right)=0\)

=>a-2b=0(Vì \(a^2+ab+3b^2=\left(a+\frac12b\right)^2+\frac{11}{4}b^2>0\forall a,b>0\)

=>a=2b

\(B=\frac{a^4-4b^4}{b^4-4a^4}\)

\(=\frac{\left(2b\right)^4-4b^4}{b^4-4\cdot\left(2b\right)^4}=\frac{16b^4-4b^4}{b^4-4\cdot16b^4}=\frac{12b^4}{b^4-64b^4}=\frac{12}{-63}=-\frac{4}{21}\)