Find all real numbers r that satisfy the equation below: $$\dfrac{1}{r^3 + 7} -7 = \frac{-r^3}{r^3 + 7}.$$

Solve for r:
1/(r^3+7)-7 = -r^3/(r^3+7)

Multiply both sides by r^3+7:
1-7 (r^3+7) = -r^3

Expand out terms of the left hand side:
-48-7 r^3 = -r^3

Add r^3+48 to both sides:
-6 r^3 = 48

Divide both sides by -6:
r^3 = -8

Taking cube roots gives 2 (-1)^(1/3) times the third roots of unity:
Answer: |  r = -2   or   r = 2 (-1)^(1/3)   or   r = -2 (-1)^(2/3)

1  /  [r^3 + 7]   - 7   =  -r^3 / [ r^3 + 7 ]    rearrange

1 / [ r^3 + 7 ] + r^3 / [ r^3 + 7]    =  7

[ r^3 + 1 ] / [ r^3 + 7 ]   = 7       cross-multiply

r^3 + 1    =   7 [ r^3 + 7 ]

r^3 + 1 =  7r^3  + 49

-6r^3   =    48     divide both sides by -6

r^3  =  -8       →   r  =  -2