Ta có \(a^{10}+b^{10}=a^{15}+b^{15}=a^{20}+b^{20}\Leftrightarrow\left(a^{15}+b^{15}\right)^2=\left(a^{10}+b^{10}\right)\left(a^{20}+b^{20}\right)\Leftrightarrow a^{30}+2a^{15}b^{15}+b^{30}=a^{30}+a^{20}b^{10}+a^{10}b^{20}+b^{30}\Leftrightarrow a^{20}b^{10}-2a^{15}b^{15}+a^{10}b^{20}=0\Leftrightarrow a^{10}b^{10}\left(a^{10}-2a^5b^5+b^{10}\right)=0\Leftrightarrow a^{10}b^{10}\left(a^5-b^5\right)^2=0\)Mà a > 0, b > 0 nên \(a^{10}b^{10}>0\) => \(\left(a^5-b^5\right)^2=0\Leftrightarrow a=b\)
Thay vào GT, ta có \(2a^{10}=2a^{15}=2a^{20}\Leftrightarrow a^5=1\Leftrightarrow a=1=b\)
Khi đó, ta có \(P=1^{2021}.2=2\)
Vậy P = 2
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