=A=\(\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\)
có (a+b)3=a3+3a2b+3ab2+b3
=a3+b3+3ab(a+b)
Ad ta có
A3=2+3(\(\sqrt[3]{1+\dfrac{\sqrt{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt{84}}{9}}\)) .
(\(\sqrt[3]{\left(1+\dfrac{\sqrt{84}}{9}\right)\left(1-\dfrac{\sqrt{84}}{9}\right)}\))
A3=2+3A\(\sqrt[3]{1-\dfrac{84}{81}}\)
A3=2-A
=>A3+A-2=0
=>A3-A+2A-2=0
=>A(A2-1)+2(A-1)=0
=>A(A-1)(A+1)+2(A-1)=0
=>(A-1)(A2+A+2)=0
=>(A-1)(A2+2.\(\dfrac{1}{2}\)A+\(\dfrac{1}{4}\)+\(\dfrac{7}{4}\))=0
=>(A-1)((A+\(\dfrac{1}{2}\))2+\(\dfrac{7}{4}\))=0
=> A=1
hoặc (A+\(\dfrac{1}{2}\))2+\(\dfrac{7}{4}\)=0(loại)
vậy A nguyên