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

 Jun 2, 2024
 #1
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\(\frac{1}{64a^3 + 7} - 7 = 0 \)

 

1/[64a^3 + 7]   =  7

 

1 =  7 [ 64a^3 + 7]

 

1 = 448a^3 + 49

 

-48  = 448 a^3

 

[ -48 /448 ] = a^3

 

-3/28 = a^3

 

a =   - ∛ ( 3/28)

 

 

cool cool cool

 Jun 2, 2024

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