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
mizoo
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)#

sankarankalyanam
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)#

Narad T.
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#

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