lokiando
  • lokiando
Which product will result in a sum or difference of cubes? a) (x + 7)(x^2 – 7x + 14) b) (x + 8)(x^2 + 8x + 64) c) (x – 9)(x^2 + 9x + 81) d) (x – 10)(x^2 – 10x + 100)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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lokiando
  • lokiando
@UsukiDoll
UsukiDoll
  • UsukiDoll
sum of cubes formula \[(a+b)(a^2-ab+b^2)\] difference of cubes formula \[(a-b)(a^2+ab+b^2)\]
lokiando
  • lokiando
c?

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More answers

lokiando
  • lokiando
will give medal
anonymous
  • anonymous
I think the answer is C.
lokiando
  • lokiando
you think? mmmmm
anonymous
  • anonymous
Not really sure on this one. I was thinking either A. or C.
UsukiDoll
  • UsukiDoll
ah I see we have to go backwards on this.. so we need all perfect cubes. . no it can't be a.
anonymous
  • anonymous
Because it can't be factored right? @UsukiDoll
UsukiDoll
  • UsukiDoll
8 27 64 \[2^3 = 8 , 3^3 =27, 4^3 = 64, 5^3 = 125\]
UsukiDoll
  • UsukiDoll
@200205650 true and also 14 is not a perfect cube
lokiando
  • lokiando
okay
lokiando
  • lokiando
so its c?
UsukiDoll
  • UsukiDoll
sum of cubes \[(a^3+b^3) = (a+b)(a^2-ab+b^2) \] difference of cubes \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \]
UsukiDoll
  • UsukiDoll
these choices are wonky though
anonymous
  • anonymous
I don't see how you can factor any of them in the answer choices...unless you use the quadratic formula.
UsukiDoll
  • UsukiDoll
like .. let a = x let b = 4 \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \] \[(x^3-4^3) = (x-4)(x^2+4x+16) \]
UsukiDoll
  • UsukiDoll
I see perfect squares for the last 3 choices that's true.. for the ending part is b^2 . It's like this question isn't following the rules. That's what's getting me mad
lokiando
  • lokiando
same here
UsukiDoll
  • UsukiDoll
let a = x and let b = 9 difference of cubes formula \[(a^3-b^3) = (a-b)(a^2+ab+b^2) \] \[(x^3-9^3) = (x-9)(x^2+9x+81) \]
anonymous
  • anonymous
From your examples so far @UsukiDoll I would think C. is the right answer.
UsukiDoll
  • UsukiDoll
that is messed up... really. 9 isn't a perfect cube to begin with. perfect square yes but not perfect cube
anonymous
  • anonymous
Ya it isn't...but the way the equations are set up B. and D. don't follow that same pattern.
UsukiDoll
  • UsukiDoll
only perfect cubes are allowed for these formulas...>:/ oh I'm missing something XD 9 x 9 x 9 = 729
UsukiDoll
  • UsukiDoll
729 is a perfect cube number. that guy is 9
UsukiDoll
  • UsukiDoll
ok I got it.. I missed a very important step let a = x and b = 9 \[(x^3-729) = (x-9)(x^2+9x+81)\] \[(x^3-9^3) = (x-9)(x^2+9x+81)\]
anonymous
  • anonymous
So the answer is C.?
UsukiDoll
  • UsukiDoll
that 729 is rewritten as 9^3 in the second line so a = x (the variable) and b = 9 (the number) a^2 = x^2 ab = 9x b^2 = (9)(9) = 81
UsukiDoll
  • UsukiDoll
yeah C matches ... on top of that .. the signs match for the difference of cubes. I wasn't working backwards until the very beginning.
anonymous
  • anonymous
Great job explaining that!
lokiando
  • lokiando
well thank you guys for your time and patients :)
UsukiDoll
  • UsukiDoll
XD patience XDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD
UsukiDoll
  • UsukiDoll
It's past my lunch time too... so that's why my thinking started to go off

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