+0

# Dealing with Cubes...

+5
82
3
+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
+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
+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
+122390
0

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)

Feb 12, 2022