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Here, we have x⁶ = (x-2)⁶ so we have following situations: (i) x=x-2 i.e. 0= -2 that is not possible (ii) both sides the power is even so x and (x-2) are additive inverse of each other. -x = x-2 -2x = -2 So x=1
nice video but I have to point it it can become confusing if you alter between x and • for multiplying. if you have algebra in the question, you should never use x for multiplying
This problem isn't stated properly. It should specify *which* solutions to find, or, equivalently, *how many* solutions to find. The way it is stated, it makes it seem like the problem is asking for a single value for x, in which case x = 1 is the most obvious and natural solution. A more appropriate way to state this problem would be "find all complex numbers x satisfying the equation x^6 = (x-2)^6". Still, nice video! :)
А если разложить как разность кубов сначала (х^2-(х-2)^2)*(х^4+(х(х-2))^4+(х-2)^4)=0 Легко доказать что второй множитель всегда больше чем 0, т.е. нет таких действительных значений х тогда остаётся только Х^2-(х-2)^2=0, что решается легко
Veel te ingewikkeld Can be solved much easier in less than a minut. X^6=(X-2)^6 X^(6/3)=(X-2)^(6/3) X^2=(x-2)^2 X^2=X^2-4X+4 0=-4X+4 4X=4 X=1 Orig.eq. 1^6=(1-2)^6 ==> 1=(-1)^6 1=1
If you are going to find complex roots, you can do it a little differently. (1) x/(x-2)=t ; (2) x=2*t/(t-1) t=NO=1; (3) x^6/(x-2)^6=t^6=1 ; (4) t^6=e^(2*pi*i*n ; n - integer . So : (5) t=e^(i*pi*n/3) . (6) n=0 : t1=e^(i*0)=e^0=1 NO !! ; n=1 : t2=e^(i*pi/3)=cos(pi/3)+i*sin(pi/3)=0,5*(1+i*sqrt(3) ) ; n=2 : t3=e^(2*i*pi/3)=…=-0,5*(1-i*sqrt(3) ) ; n=3 : t4=e^(3*i*pi/3)=-1 ; ………. n=5 : t6=e^(5*i*pi/3)=….=0,5*(1-sqrt(3) ) ; n=6 : t7=t1 ….. We substitute (6) in (2) - we get Your answer. With respect , Lidiy
Congratulations. You just refuted the fundamental theorem of algebra🤣🤣🤣. " Every, non zero, single variable, degree n polynomial with complex coefficients has, counted with mutiplicity, exactly n complex roots". You are bigger than Gauss 😂😂😂😂
