a, \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=\left(x^2y+xy^2\right)+\left(x^2z+xyz\right)+\left(xz^2+yz^2\right)+\left(yz^2+xyz\right)\)
\(=xy\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+yz\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz+yz+z^2\right)=\left(x+y\right)[x\left(y+z\right)+z\left(y+z\right)]\)
\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
b,\(x^{16}+x^8-2=x^{16}+x^8-1-1=\left(x^{16}-1\right)+\left(x^8-1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)+\left(x^8-1\right)=\left(x^8-1\right)\left(x^8+2\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+2\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)c,\(A=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)+1\)\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
Đặt \(a^2+5a+5=k\) thế vào biểu thức A ta có:
\(A=\left(k-1\right)\left(k+1\right)+1=k^2-1+1=k^2=\left(a^2+5a+5\right)^2\)
