(1)由已知a+b=6可得(a+b) 2=36,即:a 2+b 2+2ab=36,
∵ab=-27,
∴a 2+b 2=36-2×(-27)=90;
(2)由(1)可得a 2+b 2=90,
∵ab=-27,
∴a 2+b 2-ab=90+27=117;
(3)∵(a-b) 2=a 2-2ab+b 2=a 2+b 2-2ab,a 2+b 2=90,
∴a 2+b 2-2ab=90-2×(-27)=144.
(1)由已知a+b=6可得(a+b) 2=36,即:a 2+b 2+2ab=36,
∵ab=-27,
∴a 2+b 2=36-2×(-27)=90;
(2)由(1)可得a 2+b 2=90,
∵ab=-27,
∴a 2+b 2-ab=90+27=117;
(3)∵(a-b) 2=a 2-2ab+b 2=a 2+b 2-2ab,a 2+b 2=90,
∴a 2+b 2-2ab=90-2×(-27)=144.