Question 832975
That one is to the power of three?  Use Rational Roots Theorem, and check for roots:  -1, -2, -3, -4, -6, -12, 1, 2, 3, 4, 6, 12...
Only three of them at most, will work as roots.  


Any trouble, then tell.






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NOTE:  I tried -1, and 1 first, and found 1 is a root, for the binomial factor of (x-1).
This resulted in {{{x^2+x-12}}} as the quotient from the synthetic division.  That quadratic factor has roots -4 and +3, using general solution to a quadratic equation, so the binomial factors are (x+4) and (x-3).
The Factorization of your given expression is {{{highlight((x-1)(x+4)(x-3))}}}