anonymous
  • anonymous
How do you solve 8x^3-27?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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precal
  • precal
it needs to be equal to something for you to solve it
anonymous
  • anonymous
its equal to zero, sorry
lgbasallote
  • lgbasallote
convert it to the form (a^3 + b^3) = (a+b)(a^2-ab + b^2)

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More answers

lgbasallote
  • lgbasallote
(2x + 3)(4x^2 - 6x + 9)...just distribute to get the final answer
anonymous
  • anonymous
so do i set the first equation to zero and solve then use the quadratic formula to solve the second equation to find the roots?
lgbasallote
  • lgbasallote
no! distribute....
anonymous
  • anonymous
i'm lost
lgbasallote
  • lgbasallote
oh wait...i was factoring :))) I should've been solving. Sorry for confusing you. Transpose 27 to the other side. 8x^3 = 27 get the cube root 2x = 3 x = 3/2
anonymous
  • anonymous
there should be two more roots..
anonymous
  • anonymous
that's okay, thank-you
lgbasallote
  • lgbasallote
@cinar...i think this is a repetitive root
anonymous
  • anonymous
no, they are complex
anonymous
  • anonymous
\[\frac34(-1+i \sqrt{3})\quad and \quad \frac34(-1-i \sqrt{3})\quad are \,also\, roots\]

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