\(A+x^4+64\)
\(A=\left(x^2\right)^2+8^2+16x^2-16x^2\)
\(A=\left(x^2+8\right)^2-16x^2\)
\(A=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(A=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)
\(x^4+64\\ \\=x^4+64+16x^2-16x^2\\ \\=\left(x^4+16x^2+64\right)-16x^2\\ \\=\left(x^2+8\right)^2-\left(4x\right)^2\\ \\=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
theo phương pháp thêm bớt 1 hạng tử
ta có :\(^{ }x^4\)+\(64x^2\)+64-\(64x^2\)
\(\Rightarrow\)(\(x^2\)+8)\(^2\)-(8x)\(^2\)
\(\Rightarrow\)(x\(^2\)+8-8x)(x\(^2\)-8+8x)
