\(5x^2+10xy+5y^2\\ =5\left(x^2+2xy+y^2\right)\\ =5\left(x+y\right)^2\)

\(c)8x^3+12x^2y+6xy^2+y^2=(2x+y)^3 \)

\(d)-x^3+9x^2-27x+27=3^3-9x^2+27x-x^3=(3-x)^3\)

\(e)\)\(x^3+\dfrac{1}{27}=x^3+\left(\dfrac{1}{3}\right)^3=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)

\(g)\)\(\dfrac{1}{25}x^2-64y^2=\left(\dfrac{1}{5}+8y\right)\left(\dfrac{1}{5}-8y\right)\)

\(\dfrac{16+x^2}{x^2-16}=\dfrac{4^2+x^2}{x^2-4^2}=\dfrac{4^2+x^2}{\left(x-4\right)\left(x+4\right)}\)

\(\dfrac{16+x^2}{x^2-16}=\dfrac{4^2+x^2}{x^2-4^2}=\dfrac{ }{ }\)

\(\dfrac{4x+16}{x^2-16}=\dfrac{4\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{4}{x-4}\)

\(-x^2-2x+8\\ =-\left(x^2-4x-2x+8\right)\\ =-\left[x\left(x-2\right)-4\left(x-2\right)\right]\\ =-\left(x-2\right)\left(x-4\right)\)

\(8x^3+12x^2y+6xy^2+y^3\\ =\left(2x\right)^3+12x^2y+6xy^2+y^3\\ =\left(2x+y\right)^3\)

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x-3\right)^2\)

\(8x^3+12x^2+6x+1\\ =\left(2x\right)^3+12x^2+6x+1^3\\ =\left(2x-1\right)^3\)

\(e)x^4-2x^4+x^2 =x^2.x^2-2x.x^2+x^2+1 =(x^2)(x^2-2x+1) =x^2(x-1)^2 \)

\(f)27y^3-x^3 =(3y)^3-x^3 =(3y-3)(9y^2+3xy+x^2)\)