a) Ta có: A3=(3√2+√5+3√2−√5)3�3=(2+53+2−53)3
=2+√5+2−√5+3⋅3√(2+√5)(2−√5)(3√2+√5+3√2−√5)=2+5+2−5+3⋅(2+5)(2−5)3(2+53+2−53)
=4−3⋅A=4−3⋅�
⇔A3+3A−4=0⇔�3+3�−4=0
⇔A3−A+4A−4=0⇔�3−�+4�−4=0
⇔A(A−1)(A+1)+4(A−1)=0⇔�(�−1)(�+1)+4(�−1)=0
⇔(A−1)(A2+A+4)=0
\(A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot A\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\)
=>\(A^3=4-3A\)
=>\(A^3-3A-4=0\)
=>\(A\simeq2,196\)