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 Feb 24, 2021
 #1
avatar+421 
+1

Taking a b out of the numerator and denominator of the first  one, we get $\frac{4b-12}{18b^2-128}=\frac{2b-6}{9b^2-64}=\frac{2b-6}{(3b+8)(3b-8)}$.

FOr the second, the denominator factors as $3(3b+2)(7b-6)$, and $27b+18=9(3b+2)$. You can do the rest, I believe in you!!

 Feb 24, 2021
 #2
avatar+483 
0

:0 I think this is for spirit of math right

 

4b2 - 12b = 4b(b - 3)

27b + 18 = 9(3b + 2)

18b3 - 128b = 2b(9b2 - 64)

                    = 2b(3b + 8)(3b - 8)

63b2 - 282b - 216 = 3(21b2 - 94b - 72)

                             = 3(3b + 2)(7b - 36)

 

Then you put these over the fractions:

 

= 4b(b - 3) / 2b(3b + 8)(3b - 8)  -   9(3b + 2) / 3(3b + 2)(7b - 36)

= 2(b - 3) / (3b + 8)(3b - 8)  -  3 / (7b - 36)

= (2(b - 3)(7b - 36)  -  3(3b + 8)(3b - 8)) / (3b + 8)(3b - 8)(7b - 36)

= (14b2 - 114b + 216 - 27b2 + 192) / (3b + 8)(3b - 8)(7b - 36)

 

= (-13b2 - 114b + 408) / (3b + 8)(3b - 8)(7b - 36)   ; b ≠ 0 , ±8/3 ,  -2/3 , 36/7

 Feb 24, 2021
 #3
avatar+16 
0

 

thats what i got 

NerdOFMathThanks  Feb 24, 2021
edited by NerdOFMathThanks  Feb 24, 2021

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