jenkinsbunch
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- Sep 24, 2008
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(8x^3-1). I thought it would be (2x^2-1)^3, but that didn't work.
I'm having trouble factoring
- Thread starter jenkinsbunch
- Start date
Hello, jenkinsbunch!
Of course, it didn't work . . .
You're expected to recognize a difference of cubes:
. . . \(\displaystyle a^3 - b^3 \:=\
a-b)(a^2+ab+b^2)\)
\(\displaystyle \text{You have: }\;(2x)^2 - 1^3\quad\hdots\quad\text{Got it?}\)
Of course, it didn't work . . .
You're expected to recognize a difference of cubes:
. . . \(\displaystyle a^3 - b^3 \:=\
a-b)(a^2+ab+b^2)\)\(\displaystyle \text{You have: }\;(2x)^2 - 1^3\quad\hdots\quad\text{Got it?}\)
Hello,
I would like to add one piece of information if you dont' mind soroban.
The difference of cubes is:
a^3 - b^3 = (a-b)(a^2+ab+b^2) That means the when you have (b^3 = c = a perfect cube, such as -1, 1, 8, 64, etc) you can simply use that formula.
There is another formula though!
This one is:
a^3 + b^3 = (a + b)(a2 – ab + b2)
I would like to add one piece of information if you dont' mind soroban.
The difference of cubes is:
a^3 - b^3 = (a-b)(a^2+ab+b^2) That means the when you have (b^3 = c = a perfect cube, such as -1, 1, 8, 64, etc) you can simply use that formula.
There is another formula though!
This one is:
a^3 + b^3 = (a + b)(a2 – ab + b2)