\(a.x^3-6x=x^3-4^3=\left(x-4\right)\left(x^2+4x+16\right)\)
\(b.x^4+6x^3+11x^2+6x+1=x^4+6x^3+9x^2+2x^2+6x+1\)
\(=\left(x^2+3x+1\right)^2\)
\(c.x^2+3x+2=x^2+x+2x+2=x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(x+2\right)\)
\(d.x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x=y\Rightarrow y\left(y+2\right)+1=y^2+2y+1=\left(y+1\right)^2\)
Thay \(y=x^2+3x\) ta được: \(\left(y+1\right)^2=\left(x^2+3x+1\right)^2\)
\(e.x^3+9x^2+27x+27=\left(x+3\right)^3\)
\(f.\left(x+1\right)\left(x+7\right)\left(x^2+8x+15\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(a=x^2+8x+11\Rightarrow\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a+1\right)\left(a-1\right)\)
Thay \(a=x^2+8x+11\) ta được: \(\left(a+1\right)\left(a-1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
