Poland | A Nice Algebra Problem

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[@SidneiMV]

x + 3 = u => x = u - 3
(u - 3)(u - 1)(u + 1)(u + 3) = 9
(u² - 1)(u² - 9) = 9
u⁴ - 10u² = 0
u²(u² - 10) = 0
u = 0 => *x = -3*
u² - 10 = 0 => u = ± √10
x = *-3 ± √10*

[@pwmiles56]

It's much easier if you make the substitution
u = x+3
(u-3)(u-1)(u+1)(u+3) = 9
(u^2-9)(u^2-1) = 9
u^4 - 10u^2 + 9 = 9
u^2(u^2 - 10) = 0
u = 0, x = -3
u = +/- sqrt(10), x = -3 +/- sqrt(10)

[@Patrik6920]

Very nice solution, not sure avery one can follow it.. but its very neat....
maby som clairifications r in order for ppl?...
..the first x term for example, proably not abvious to everyone...
..and the u²(u²-10)=0 wich has 4 roots of wich two are 0 or ±0=0

[@pwmiles56]

@@Patrik6920 Hi Patrik, good points. The idea at work was to change the equation to make it more symmetrical. My "u" is arranged to be in the middle of the bracketed terms i.e. half-way between x and x+6. A strategy like this often pays off, but I held off from an exposition dump, as it probably wouldn't have helped. Thanks again.

[@Patrik6920]

@@pwmiles56 np, good solution...