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How do you factor #x^6-7x^3-8#?

1 Answer

Apr 18, 2015

Like this,

#x^6-7x^3-8 -= (x^3)^2 -7(x^3) -8 -= (x^3-8)(x^3+1) -= (x^3-(2)^3)(x^3+(1)^3)#

Recall that, #a^3+b^3=(a+b)(a^2-ab+b^2)# and # a^3-b^3=(a-b)(a^2+ab+b^2)#

#=> (x-2)(x^2+2x+4)(x+1)(x^2-x+1)#

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