A Mixed Radical

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[@Qermaq]

This was a fun one to solve. I used u^6 = x, so u^3 = sqrt(x) and u^2 = cbrt(x). Wound up with a quartic 3u^4 - 2u^2+3u^2=0. Factored out u^2 (x = 0 is an obvious solution) and we're left with a quadratic where u = 1/3 +/- (2sqrt(2)/3)i. Raising this to the 6th power was the worst part, but I got x = 329/729 +/- (460sqrt(2)/729)i. SO three distinct solutions (because 0 came up twice). Crossing my fingers on this one. EDIT: yeah, I double-checked my complex solution and it's messed up. Not sure why. Oh well my brain is tired.

[@phill3986]

😊😊😊👍👍👍

[@lawrencejelsma8118]

Something went wrong with your extended analysis. I thought x = 1 is also a solution and your Wolfram Alpha graphs must be based on the remainder polynomial of x=0 as the only other real solution!🤣
At 3:24 you were first talking about a try solution of y^2 = 1 and y = +/- 1 from that gave the y = 1 and the extraneous other solution y = -1 other solution.
I think you should have discussed factoring your first x = y^6 - 2y^3 + 1 into (y - 1) ( quadratic of a square remainder) to know y = 1 solves before the 3:24 mark)!!