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avatar+251 

Find all values of a that satisfy the equation \(\frac{1}{a^3 + 7} -7 = \frac{-a^3}{a^3 + 7}.\)

 

I multiplied both sides by a^3+7 to get -6=-a^3.

 

I think a solution to a would be \(a=\sqrt[3]{6}\) but I'm not sure...

 Feb 12, 2022
 #1
avatar+1209 
+2

That's a good step, but the problem with your reasoning is that -7 by itself doesn't have a^3 + 7 in the denominator, so you would have to convert -7 to (-7(a^3 + 7))/(a^3 + 7) before solving.

 Feb 12, 2022
 #2
avatar+251 
+4

Whoops, I didn't notice that. Thanks for catching it!

Correcting the mistake I think it would be \(a=\sqrt[3]{-8}\)

dolphinia  Feb 12, 2022
 #3
avatar+122390 
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Rewrite as  :

 

1/ (a^3 + 7)  +  a^3 / (a^3 + 7)   =   7

 

(1 + a^3 ) /  (a^3 + 7)   =  7

 

Multiply both sides  by a^3 + 7

 

1 + a^3   = 7 ( a^3 + 7)

 

1 + a^3  = 7a^3  + 49

 

1 - 49  = 7a^3  - a^3

 

-48  =  6a^3

 

-8  = a^3       take the cube root

 

-2  = a    ( only real solution)

 

 

 

cool cool cool

 Feb 12, 2022

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