How do you factor $(x^2-9)×(x^3-8)$ ?

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Answer 1 · 2015-04-09T01:07:41

$(x^2-9) = (x-2)(x+2)$

$(x^3-8) = (x-2)(x^2+2x+4)$
(note that $(x^2+2x+4)$ has no Real factors; this may have been an intentional trick)

So
$(x^2-9)xx(x^3-8)$
can be factored as
$(x-2)^2(x+2)(x^2+2x+4)$

How do you factor (x^2-9)×(x^3-8)(x^2-9)×(x^3-8)?