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Find all real numbers a that satisfy \frac{1}{64a^3 + 7} - 7 = 0.

 Jun 22, 2024
 #1
avatar+129881 
+1

1  / (64a^3 + 7)  - 7 =  0

 

1 / (64a^3 + 7)  =  7

 

1 = 7 (64a^3 + 7)

 

1 = 448a^3 + 49

 

-48 = 448a^3

 

-48 / 448  = a^3

 

-3/28 = a^3

 

a = (-3/28)^(1/3)

 

 

cool cool cool

 Jun 22, 2024

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