Có \(a^{12}+b^{12}=a^{12}+a^{11}b-a^{11}b+ab^{11}-ab^{11}+b^{12}\)
\(=a^{11}\left(a+b\right)+b^{11}\left(a+b\right)-a^{11}b-ab^{11}\)
\(=\left(a^{11}+b^{11}\right)\left(a+b\right)-ab\left(a^{10}+b^{10}\right)\)
\(=\left(a^{12}+b^{12}\right)\left(a+b\right)-ab\left(a^{12}+b^{12}\right)\)(vì giả thiết cho \(a^{10}+b^{10}=a^{11}+b^{11}=a^{12}+b^{12}\))
\(=\left(a^{12}+b^{12}\right)\left(a+b-ab\right)\)
Đã chứng minh \(a^{12}+b^{12}=\left(a^{12}+b^{12}\right)\left(a+b-ab\right)\)suy ra:
\(a+b-ab=1\)
=> \(a+b-ab-1=0\)
=> \(a-1-b\left(a-1\right)=0\)
=> \(\left(a-1\right)\left(1-b\right)=0\)
=> \(a=1\)hoặc \(b=1\)
Nếu \(a=1\)thì từ giả thiết suy ra
\(b^{10}+1=b^{11}+1\)
=> \(b^{10}=b^{11}\)suy ra \(b^{10}\left(b-1\right)=b^{11}-b^{10}=0\)
Mà đề cho b dương =>\(b=1\)=>\(P=a^{20}+b^{20}=2\)
Nếu \(b=1\)thì từ giả thiết suy ra
\(a^{10}+1=a^{11}+1\)
=> \(a^{10}=a^{11}\)suy ra \(a^{10}\left(a-1\right)=a^{11}-a^{10}=0\)
Mà đề cho a dương =>\(a=1\)=>\(P=a^{20}+b^{20}=2\)