\(x^{12}+4=\left(x^6\right)^2+4=\left(x^6\right)^2+4x^6+4-4x^6\)
\(=\left(x^6+2\right)^2-\left(2x^3\right)^2\)
\(=\left(x^6+2+2x^3\right)\left(x^6+2+2x^3\right)\)
Ta có: \(x^{12}+4\)
\(=x^{12}+4x^6+4-4x^6\)
\(=\left(x^6+2\right)^2-\left(2x^3\right)^2\) (hằng đẳng thức)
\(=\left(x^6+2+2x^3\right)\left(x^6+2+2x^3\right)\)
\(=\left(x^6+2+2x^3\right)^2\)
\(x^{12}+4=\left(x^6\right)^2+4\)
\(=\left(x^6\right)^2+4x^6+4-4x^6\)
\(=\left(x^6+2\right)^2-\left(2x^3\right)^2\)
\(=\left(x^6+2+2x^3\right)\left(x^6+2-2x^3\right)\)
Rảnh quớ nên làm lại cái nỳ!!!!