anonymous
  • anonymous
Can someone please factor out (x + 3) cubed?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jim_thompson5910
  • jim_thompson5910
it's already factored
jim_thompson5910
  • jim_thompson5910
did you mean expand?
anonymous
  • anonymous
Perhaps I mean expand

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anonymous
  • anonymous
The other issue is expanding (x + delta x) cubed?
jim_thompson5910
  • jim_thompson5910
Ok first off we need to expand out (x + 3)(x + 3) (x + 3)(x + 3) = x(x+3) + 3(x+3) (x + 3)(x + 3) = x^2 + 3x + 3x + 9 (x + 3)(x + 3) = x^2 + 6x + 9
jim_thompson5910
  • jim_thompson5910
then we use this to expand out (x + 3)(x + 3)(x + 3)
jim_thompson5910
  • jim_thompson5910
(x + 3)^3 = (x + 3)(x + 3)(x + 3) (x + 3)^3 = (x + 3)(x^2 + 6x + 9) (x + 3)^3 = x(x^2 + 6x + 9) + 3(x^2 + 6x + 9) (x + 3)^3 = x^3 + 6x^2 + 9x + 3x^2 + 18x + 27 (x + 3)^3 = x^3 + 9x^2 + 27x + 27
jim_thompson5910
  • jim_thompson5910
you expand (x + delta x) cubed the same way as well
anonymous
  • anonymous
I seem to be making an algebraic error in expanding (x + delta x) cubed. Could you please work it out for me. Thank you so much.
jim_thompson5910
  • jim_thompson5910
replace delta x with another variable (say y) and try it again
anonymous
  • anonymous
\[(a+b)³=a^3+3a²b+3ab²+b³\] Just change a to x and b to delta x.
anonymous
  • anonymous
Well, the problem you see is how Delta X is treated in a multiplicative relation with plain X or X squared.
jim_thompson5910
  • jim_thompson5910
not sure what you mean rafirank
anonymous
  • anonymous
Jim--thank you and thank you very much to Zairhenrique. I've got it now. Thanks to you both. I understand what I have been doing wrong.
jim_thompson5910
  • jim_thompson5910
ok glad it's making sense now
anonymous
  • anonymous
you welcome man

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