A Nice math Olympiad exponential problem.

[@sylvesterogbolu-otutu1498]

But x^3 = 3^6, x = 9 is just one solution in N. There should be two other solutions in R. x^3 = 3^6 => x^3 = 9^3 x^3 - 9^3 = 0 (Transpose set to 0) (x - 9)(x^2 + 9x + 9) = 0 (Difference of two cubes) x - 9 = 0; x = 9 € N. Solving x^2 + 9x + 9 = 0 using the quadratic formula will give two irrational roots in R. (- 9 +/- 3(sqrt5))/2.

[@ganeshdas3174]

x.(x)^1/2 = 27 x^(1+1/2) =3^3 x^3/2 =3^3 (x^3/2)^2/3 = (3^3)^2/3 x = 3^2 = 9

[@JJONLINEMATHSCLASSchannel]

👏👏👏

[@amtrakatsfnyc]

This equation generates a cubic equation which has 3 roots. The 3 roots are 9 and 2 complex numbers which are conjugates of each other.

[@abrahammg9814]

X=9

[@AliHassan-hb1bn]

Radical equation has only one solution.

[@AliHassan-hb1bn]

Ra

[@Extratremendouszeus]

But you are talented...teach me to be as super as you are darling😊