`1)`\(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]\)
`2)`\(x^3\left(y-1\right)^3=\left[x\left(y-1\right)\right]^3\)
`3)`\(125-\left(x+2\right)^3=5^3-\left(x+2\right)^3=\left(5-x-2\right)\left[5^2+5\left(x+2\right)+\left(x+2\right)^2\right]=\left(3-x\right)\left(25+5x+10+x^2+4x+4\right)=\left(3-x\right)\left(x^2+9x+39\right)\)
`4)`\(\left(x+3\right)^3-8=\left(x+3\right)^3-2^3=\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+2^2\right]=\left(x+1\right)\left(x^2+6x+9+2x+6+4\right)=\left(x+1\right)\left(x^2+8x+19\right)\)
`5)`\(\left(x-5\right)^3-27=\left(x-5\right)^3-3^3=\left(x-5-3\right)\left[\left(x-5\right)^2+3\left(x-5\right)+3^2\right]=\left(x-8\right)\left(x^2-10x+25+3x-15+9\right)=\left(x-8\right)\left(x^2-7x+19\right)\)
