已知a-b=根号3+根号2,b-c=根号3-根号2,求a平方+b平方+c平方-ab-bc-ca

a-b=√3+√2b-c=√3-√2a-c=(a-b)+(b-c)=(√3+√2)+(√3-√2)=2√3a²+b²+c²-ab-bc-ca=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)]=[(a-b)²+(b-c)²+(c-a)²]/2 =[(√3+√2)²+(√3-√2)²+(2√3)²]/2 =[3+2+2√6+3+2-2√6+12]/2 =22/2 =11