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How do you factor completely $8 x^{3} - 8$ $8x^3-8$ ?

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Konstantinos Michailidis

Nov 16, 2015

It is

$8 x^{3} - 8 = 8 \cdot (x^{3} - 1) = 8 \cdot (x - 1) \cdot (x^{2} + x + 1)$ $8x^3-8=8*(x^3-1)=8*(x-1)*(x^2+x+1)$

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