What concentration of Ag+ is in equilibrium with 3.0×10−6 M Co(CN)3−6 and Ag3Co(CN)6(s)?

Ag3Co(CN)6 --> 3Ag^+3 + Co(CN)6 ^-3
I worked the problem as such:
Ksp= 3.9*10^-26
3.9*10^-26= [3x]^3 * [x]
3.9*10^-26= [3x]^3 * [3.0*10^-6]
Divided both sides by 3.0*10^-6 to isolate the [3x]^3, which yields:
1.3*10^-20=[3x]^3

Divided both sides by 3 to isolate [x]^3, which yields:
4.33*10^-21=[x]^3

Took the cube root of both sides to isolate x, which yields:
1.63*10^-7=x

Yet this answer is apparently wrong. Can you help me?

3 answers

1.3*10^-20=[3x]^3

Divided both sides by 3 to isolate [x]^3, which yields:
4.33*10^-21=[x]^3
See your work above. That term of (3X)^3 is 27X^3 so the next step is to divide by 27 and not 3. I didn't check anything after that.
Thank you. Unfortunately I still got an incorrect answer after doing:
1.3*10^-20=[3x]^3 divide both sides by 27, which yields:

4.81*10^-22=[x]^3 then I took the cubed root (3√(4.81*10^-22)
Which yields: 7.835*10^-8=x

Thank you very much for your time and help! I appreciate it so much.
It may be worthwhile to check your answer again with significant figures. The use of 3.9E-26 limits the number of s.f. to 2 so
the answer should be reported as 7.8E-8 M. Some online databases will count too many significant figures as an incorrect answer.