How do you find x?

x+(1/x^3)=0

I know the answer is
1-(3/x^4)
but I can't found why

3 Answers
Jul 2, 2018

See Expectation please.

Explanation:

x + 1/x^3 = 0

(x^4 + 1)/x^3 = 0

x^4 + 1 = 0

(x^2 + sqrt -1)(x^2 - sqrt -1) = 0

(x^2 + sqrt -1) (x - sqrt sqrt -1)(x + sqrt sqrt -1) = 0

x = +- sqrt sqrt -1, x = (-1)^(1/4), x = -(-1)^(1/4)

x = sqrt (-i), -sqrt (-i) , x = (-1)^(1/4), x = -(-1)^(1/4)

Jul 2, 2018

color(maroon)(x = +- sqrt (+-i)

Explanation:

x + 1/ x^3 = 0

(x^4 + 1) / x^3 = 0

x^4 + 1 = 0

x^4 = -1

x^2 =+- sqrt (-1) = +- i

color(maroon)(x = +- sqrt(+- i)

Jul 2, 2018

The solutions are S={sqrt2/2+isqrt2/2,-sqrt2/2+isqrt2/2, -sqrt2/2-isqrt2/2, sqrt2/2-isqrt2/2 }

Explanation:

x+1/x^3=0

(x^4+1)/x^3=0

x^4+1=0

x^4=-1

Change to the polar form

x^4=cos(pi)+isinpi

x^4=e^(i(pi+2kpi)), k in ZZ

x=e^(i(pi/4+kpi/2))

When k=0

x_0=e^(ipi/4)=cos(pi/4)+isin(pi/4)=sqrt2/2+isqrt2/2

When k=1

x_1=e^(i(pi/4+pi/2))=cos(3pi/4)+isin(3pi/4)=-sqrt2/2+isqrt2/2

When k=2

x_2=e^(i(pi/4+pi))=cos(5pi/4)+isin(5pi/4)=-sqrt2/2-isqrt2/2

When k=3

x_3=e^(i(pi/4+3pi/2))=cos(7pi/4)+isin(7pi/4)=sqrt2/2-isqrt2/2