a)\(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(=\left(x^2+5xy+6y^2\right)\left(x^2+5xy+4y^2\right)+y^4\)

Let \(x^2+5xy=a;y^2=b\) we have;

\(=\left(a+6b\right)\left(a+4b\right)+b^2\)

\(=a^2+10ab+24b^2+b^2\)

\(=a^2+10ab+25b^2=\left(a+5b\right)^2\)

\(=\left(x^2+5xy+5y^2\right)^2\)

b)\(B=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+3x+2\right)\left(x^2+3x\right)+1\)

Let \(a=x^2+3x\) we have:

\(=a\left(a+2\right)+1=a^2+2a+1\)

\(=\left(a+1\right)^2=\left(x^2+3x+1\right)^2\)

Done !!

C = 4x(x + y)(x + y + z)(x + z) + y²z²

   = 4[x(x + y + z)][(x + y)(x + z)] + y²z²

   = 4(x² + xy + xz)(x² + xy + xz + yz) + y²z²

   = 4(x² + xy + xz)² + 4yz(x² + xy + xz) + y²z²

   = [2(x² + xy + xz)]² + 2.2(x² + xy + xz).yz + (yz)²

   = [2(x² + xy + xz) + yz)]²

We have things to prove