Question 156799
Yes.
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After pulling out the 3, the remainder is a "difference of cubes" -- now, you can apply "special factoring".
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See:
http://www.purplemath.com/modules/specfact2.htm
for detailed information
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3(x^3-8)
3(x-2)(x^2 + 2x + 2^2)
3(x-2)(x^2 + 2x + 4)