+) Cách $1$:
`\qquad x^3-7x-6``=x^3-x-6x-6`
`=x(x^2-1)-6(x+1)``=x(x-1)(x+1)-6(x+1)`
`=(x+1)(x^2-x-6)=(x+1)(x^2-3x+2x-6)`
`=(x+1).[x(x-3)+2(x-3)]`
`=(x+1)(x+2)(x-3)`
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+) Cách $2$:
`\qquad x^3-7x-6``=x^3-4x-3x-6`
`=x(x^2-4)-3(x+2)``=x(x-2)(x+2)-3(x+2)`
`=(x+2)(x^2-2x-3)=(x+2)(x^2-3x+x-3)`
`=(x+2)[x(x-3)+x-3]=(x+2)(x-3)(x+1)`
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+) Cách $3$:
`\qquad x^3-7x-6``=x^3-3^3-7x+21`
`=(x-3)(x^2+3x+9)-7(x-3)`
`=(x-3)(x^2+3x+9-7)``=(x-3)(x^2+x+2x+2)`
`=(x-3)[x(x+1)+2(x+1)]=(x-3)(x+1)(x+2)`
$\\$
Vậy `x^3-7x+6=(x+1)(x+2)(x-3)`
