I substituted a=x3 which gave me a^2 -a -2 = 0, solving for a you get (a-2)(a+1)=0 so a=2 or a=-1 substitue x^3 for a and you get directly to x=2^(1/3) or x=-1
Yes, there are 6 roots, two of which are real, and four are complex. It is obvious that x^6-x^3-2=(x+1)(x^3-2)(x^2-x+1): the first factor from the observation that x=-1 is a solution, and that if x^3=2 then x is a solution. The third factor can be obtained by dividing x^6-x^3-2 by the first two factors. From the first factor we have the solutions -1, from the second factors we have 2^(1/3) and the two complex solutions -2**(1/3)/2 +2^**(1/3)*sqrt(3)*i/2 and -2**(1/3)/2 - 2**(1/3)*sqrt(3)*i/2, and finally x=1/2 + sqrt(3)*i/2 and x=1/2 - sqrt(3)*i/2 from the third factor. This is all really very straightforward.
Very nice sum
substraction* (unless you're counting the rule a - b = a + (-b))
I substituted a=x3 which gave me a^2 -a -2 = 0, solving for a you get (a-2)(a+1)=0 so a=2 or a=-1 substitue x^3 for a and you get directly to x=2^(1/3) or x=-1
Thank you.
Good solution ❤
Professor, não seria mais fácil descobrir o valor do logaritmo X ao cubo = Y calculando a equação de 2° grau utilizando a fórmula de Bhaskara?
Baskhara, logaritmo e equação do 2ndo grau?😢😅😮
Оно же в уме решается😂
2^(1/3)😅😅😅
Shouldn’t there be a total of six roots. You only have 2 real plus 2 complex.
Yes, there are 6 roots, two of which are real, and four are complex. It is obvious that
x^6-x^3-2=(x+1)(x^3-2)(x^2-x+1): the first factor from the observation that x=-1 is a solution, and that if x^3=2 then x is a solution. The third factor can be obtained by dividing x^6-x^3-2 by the first two factors. From the first factor we have the solutions -1, from the second factors we have 2^(1/3) and the two complex solutions -2**(1/3)/2 +2^**(1/3)*sqrt(3)*i/2 and -2**(1/3)/2 - 2**(1/3)*sqrt(3)*i/2, and finally x=1/2 + sqrt(3)*i/2 and x=1/2 - sqrt(3)*i/2 from the third factor. This is all really very straightforward.
X = 1 is also possible
X=1 is not a solution.
Excelente solución, me gusta el estilo... Algo así como ua-cam.com/video/mB8gGT2OYmw/v-deo.html