\(\cdot a^2+b^2=2\left(8+ab\right)\)
⇔\(a^2+b^2=16+2ab\)
⇔\(\left(a-b\right)^2=16\)
mà a < b
⇒\(a-b=-4\)
\(\cdot P=a^2\left(a+1\right)-b^2\left(b-1\right)+ab-3ab\left(a-b+1\right)+64\)
\(=\left(a^3-b^3\right)+a^2+b^2+ab-3ab\left(-3\right)+64\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+a^2+b^2+10ab+64\)
\(=-4a^2-4ab-4b^2+a^2+b^2+10ab+64\)
\(=-3a^2-3b^2+6ab+64\)
\(=-3\left(a^2-ab+b^2\right)+64\)
\(=-3\left(a-b\right)^2+64\)
\(=-48+64=16\)