anonymous
  • anonymous
What are the real or imaginary solutions of the polynomial equation. 64x^3+27=0 Please help!!!! medal and fann!!!!
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
do you know how to factor the sum of two cubes?
anonymous
  • anonymous
no i cant say i do.
anonymous
  • anonymous
i skipped 3 units so idk how to do any of this?

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anonymous
  • anonymous
ok i will show you \[a^3+b^3=(a+b)(a^2-ab+b^2)\]
anonymous
  • anonymous
in this case you have \[64x^3+27=(4x)^3+3^3\] put \(a=4x,b=3\) and plug them in
anonymous
  • anonymous
so 4x^3+3^3=(4x+3)(4x^2-4x*3+3^2) right?
anonymous
  • anonymous
is that what you meant by plug them in? cause that was a formula right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
oops no
anonymous
  • anonymous
but close
anonymous
  • anonymous
\[4x^3+3^3=(4x+3)(4x^2-4x*3+3^2)\] is what you wrote but you need to square the 4 as well, should be \[4x^3+3^3=(4x+3)(16x^2-12x+9)\]
anonymous
  • anonymous
oh ok that makes more sense
anonymous
  • anonymous
what do i do from there?
geerky42
  • geerky42
\[0 = (4x)^3+3^3=(4x+3)(16x^2-12x+9)\] So either \(4x+3 = 0\) OR \(0 = 16x^2-12x+9\) So from here, you can easily solve first equation, then for second equation, just use quadratic formula.

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