\(1,\left(x^2-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]\left[\left(x+1\right)\left(x^2-x+1\right)\right]=\left(x^3-1\right)\left(x^3+1\right)=x^6-1\)
\(2:\text{tự tính}:\left(x^2-2x+4\right)\left(x+2\right)=x^3+8;\left(x+1\right)^3=x^3+3x^2+3x+1;3\left(x+1\right)\left(x-1\right)=3x^2-3\)
\(3:\left(5x-2\right)^2+\left(3x-2\right)^2-2.\left(2-5x\right)\left(2-3x\right)=\left(2-5x\right)^2+\left(2-3x\right)^2-2\left(2-5x\right)\left(2-3x\right)=\left(-2x\right)^2=4x^2\)