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Find all real numbers \(a\) that satisfy:

\([\frac{1}{a^3 + 7} -7 = \frac{-a^3}{a^3 + 7}]\)

 Jul 24, 2022
 #1
avatar+124676 
+3

1/ ( a^3 + 7)   - 7 =  -a^3 / ( a^3 + 7)      rearrange as

 

1 / (a^3 + 7)  + a^3/ (a^3 + 7)  =  7

 

(1 + a^3)  = 7 ( a^3 + 7)

 

1 + a^3   7a^3 + 49

 

-6a^3  = 48

 

a^3  =  -48 / 6

 

a^3 =  -8

 

a  = -2

 

 

cool cool cool

 Jul 24, 2022
 #2
avatar+11 
-2

1/ ( a^3 + 7)   - 7 =  -a^3 / ( a^3 + 7)      rearrange as

 

1 / (a^3 + 7)  + a^3/ (a^3 + 7)  =  7

 

(1 + a^3)  = 7 ( a^3 + 7)

 

1 + a^3   7a^3 + 49

 

-6a^3  = 48

 

a^3  =  -48 / 6

 

a^3 =  -8

 

a  = -2

 Jul 24, 2022
 #3
avatar+51 
0

Whats the point of copying CPhill's answer? 

There is no need for the same work twice. How about next time do it yourself? :)

Limpeklimpe  Jul 27, 2022

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