What is x if (7x + 1)^1/3 - x = 1 ?

To find x if (7x+1)^(1/3) - x =
1.

Substitute  7x+1 = t^3,  x = (t^3-1)/7 in the given
equation :

t - (t^3-1)/7 =
1

7t -t^3 +1  = 7

7t -t^3 +1-
7 = 0

7t-t^3 -6 = 0

Multiply
by -1:

t^3 -7t + 6 = 0 which is satisfies for t =1 as
1^3-7*1+6 = 0.

Therefore , t-1
factor.

Similarly if t= 2 also satisfy t^2-7t+6 = 0, as
2^3-2*7+6 = 0.

Also  for t = -3,  t^3-7t+6 = (-3)^3-7(-3)+6
= -27+21+6 = 0.

t =1 gives x = (t^3-1)/7 = (1-1)/7 =
0

t = 2 gives x = (2^3-1)/7 =
1

t = -3 gives x = (-3^3-1)/7 =
-4.

Thereore x = 0, 1 and
-4.

Verification: Put x= 0 in (7x+1)^(1/3) -x =
(7*0+1)^(1/3)-0 =1-0 = 1, verifies with RHS.

Put x = 1 in
(7x+1)^(1/3) - x = (7*1+1)^(1/3)-1 = 2-1 = 1 verifies
RHS.

Put x = -4 in (7x+1)^(1/3) - x = (7*(-4) +1)^(1/3) -
(-4) = (-27)^(1/3) - (-4) = -3+4 = 1 verifies with RHS.