There are 6 components (3 pairs) and one of components is 1; IF x is odd and < 0 than SUM of each of pairs consists of EVEN and ODD components equal to 0. If x=-1 the sum of each pair =0

complex 6th roots of unity

I was originally thinking of factoring and grouping and came out with a mess. Then I paid attention to the fact that not only was the LHS equal to 0, there were three even powers and three ofd powers. From there the solution became easy to find. Took me less than 15 seconds, once I spotted the pattern.

@user-kk4jv8ke1y

14 күн бұрын

Очень нужно решает

@user-kk4jv8ke1y

14 күн бұрын

Нудно

Hocam x³+x²+x -16=0 denkleminde çözüm kümesi nedir?

@onkotonkoblu

19 күн бұрын

Uhhh, 7 maybe?

x^3(x^2+x+1)+x^2+x+1=0 (x^2+x+1)(x^3+1)=0 (x^2+x+1)(x+1)(x^2-x+1)=0 ...

x^4(x+1)+x^2(x+1)+x+1=0 (x+1)(x^4+x^2+1)=0 x+1=0 x=-1 x^4+x^2+1=0 x^2=y y^2+y+1=0 y=[-1+-rq(1-4)]/2 non reale