\( a){x^3} + 3{x^2} + 3x + 1\\ = {x^3} + 3.{x^2}.1 + 3.x{.1^2} + {1^3}\\ = {\left( {x + 1} \right)^3}\\ b){x^3} - 6{x^2} + 12x - 8\\ = {x^3} - 3.{x^2}.2 + 3.x{.2^2} - {2^3}\\ = {\left( {x - 2} \right)^3}\\ c){x^2} - 2xy + {y^2} - 16\\ = {\left( {x - y} \right)^2} - 16\\ = \left( {x - y - 4} \right)\left( {x - y + 4} \right)\\ d)49 - {x^2} + 2xy - {y^2}\\ = 49 - \left( {{x^2} - 2xy + {y^2}} \right)\\ = 49 - {\left( {x - y} \right)^2}\\ = \left[ {7 - \left( {x - y} \right)} \right]\left[ {7 + \left( {x - y} \right)} \right]\\ = \left( {7 - x + y} \right)\left( {7 + x - y} \right) \)
\(x^3+3x^2+3x+1=\left(x^3+1\right)+3\left(x^2+x\right)=\left(x+1\right)\left(x^2-x+1\right)+3x\left(x+1\right)=\left(x+1\right)\left(x^2-x+3x+1\right)=\left(x+1\right)\left(x^2+2x+1\right)=\left(x+1\right)\left(x+1\right)^2=\left(x+1\right)^3\)
\(x^3-6x^2+12x-8=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)=x^2\left(x-2\right)-4x\left(x-2\right)+4\left(x-2\right)=\left(x-2\right)\left(x^2-4x+4\right)=\left(x-2\right)\left(x-2\right)^2=\left(x-2\right)^3\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
\(49-x^2+2xy-y^2=49-\left(x-y\right)^2=\left(7-x+y\right)\left(7+x-y\right)\)
\(x^2+7x+12=x^2+3x+4x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+4\right)\left(x+3\right)\)
