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9K^3-9m^2K ------------ is equivalent to 3k+ (?) 3K^2-3mK
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(A) m (B) 3m (C) 1 (D) 2 (E) 3
\[9k^3-9m^2k=9k(k^2-m^2)=9k(k-m)(k+m)\]
Do I distribute? @Mertsj
Factor the denominator and cancel any common factors.
3k(k^3-m) ...?
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\[3k^2-3mk=3k(k-m)\]
how did you get rid of the ^3?
The denominator does not have an exponent 3
oh..
\[\frac{9k^3-9m^2k}{3k^2-mk}=\frac{9k(k-m)(k+m)}{3k(k-m)}=3(k+m)\]
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Is it B?
yes
thank you!
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