Determine the three cubic roots of 8. Give the answer in the form a + bj. Also give the main root in the exponential form.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Supposing that you should solve the equation `z^3 + 8 = 0` , you need to do the following steps such that:

`z^3 + 8 = 0 => z^3 + 2^3 = 0`

You need to use the following formula such that:

`a^3 + b^3 = (a + b)(a^2 - ab + b^2)`

`z^3 + 2^3 = (z + 2)(z^2- 2z + 4)`

Since  `z^3 + 2^3 = 0 => (z + 2)(z^2 - 2z + 4) = 0`

`z = -2`

`z^2 - 2z + 4 = 0`

Using quadratic formula yields:

`z_(1,2) = (2+-sqrt(4 - 16))/2 => z_(1,2) = (2+-sqrt(-12))/2`

`z_(1,2) = (2+-2i*sqrt3)/2`  (use complex number theory `sqrt(-1) = i` )

`z_1 = 1 + isqrt3 ; z_2 = 1 - isqrt3`

Hence, evaluating the solutions to the given equation yields `z = -2 ; z = 1 + isqrt3 ; z = 1- isqrt3.`

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neela | High School Teacher | (Level 3) Valedictorian

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Let the root of 8 be x.

=> x^3=8

=> x^3-8=0

=> x^3-2^3=0

=>(x-2)(x^2+x*2+2^2)=0

=>(x-2)=0 Or x^2+2x+4=0

=> x=2 is the real cube root of 8.

Or x = {-2 +or- sqrt((2)^2-4*4)}/2x = 2{-1 +or- sqrt(-3)}/2 => x={-1+ sqrt(-1) sqrt3} Or  x=-{1+ sqrt(-1) sqrt3} are the pair of complex cube roots of 8.

 

We know cost+i*sint = e^(i*t)

 

So the complex cube root, {-1+ sqrt(-1) sqrt3} could be written as

-1/2+ i* (sqrt3)/2 = cos(2p/3)+isin(2p/3) = e^(i*2p/3), where i = sqrt(-1).

Similarly the other root  x=-{1+ sqrt(-1) sqrt3} = cos(4p/3)+i*sin(4p/3) = e^(i*4p/3).

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