ta có: \(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2=\left(a+b+c\right)^2+\left(a+b-c-2c\right)\left(a+b-c+2c\right).\)
\(=\left(a+b+c\right)^2+\left(a+b-3c\right)\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
(a+b+c)^2+(a+b-c)^2-4c^2
=(a^2+b^2+c^2+2ab+2bc+2ac)+(a^2-2ab+b^2-2ac+c^2-abc)-4c^2
=a^2+b^2+c^2+2ab+2bc+2ac+a^2-2ab+b^2-2ac+c^2-abc-4c^2
=(a^2+a^2)+(b^2+b^2)+(c^2+c^2)+(2ab-2ab)+(2bc-2bc)+(2ac-2ac)-4c^2
=2a^2+2b^2+2c^2-4c^2
=(2a^2+2b^2)+(2c^2-4c^2)
=2*(a^2+b^2)+2c^2*(1-2)