[질문] a제곱+b제곱+c제곱-ab-bc-ac (a-b=5) (b-c=4) 구하는공식이랑 답좀

a제곱+b제곱+c제곱-ab-bc-ac=1/2{(a-b)^2+(b-c)^2+(c-a)^2}=1/2(25+16+81)=61 a-b(5)+b-c(4)=a-c=9, (a-c)^2=(c-a)^2=81