a, Ta có:

\(VP=\left(a+b\right)^3-3ab\left(a+b\right)\\ =a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\\ =a^3+b^3=VT\)

Vậy..............(đpcm)

b, Ta có:

\(VT=a^3+b^3+c^3-3abc\\ =\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\\ =\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b\right)-3abc\\ =\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b+c\right)\\ =\left(a+b+c\right)\left[\left(a+b+c\right)^2-3c\left(a+b\right)-3ab\right]\\ =\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab\right)\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\)

Vậy..............(đpcm)

\(A=\left(x^2+3x+4\right)^2=\left(x^2+3x+2,25+1,75\right)^2\\ =\left[\left(x+1,5\right)^2+1,75\right]^2\ge3,0625\)

Dấu "=" sảy ra khi và chỉ khi \(x=-1,5\)

Vậy \(A_{min}=3,0625\) đạt được khi và chỉ khi \(x=-1,5\)

S = 6 x 3,14 = 18.84

hoặc lấy 6 : 2 = 3

              3 x 2 x 3,14 = 18,84