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$\exponential{(x)}{3} + 8 $
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\left(x+2\right)\left(x^{2}-2x+4\right)
Rewrite x^{3}+8 as x^{3}+2^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right). Polynomial x^{2}-2x+4 is not factored since it does not have any rational roots.