Find all solutions of the equation 213-8p3=0

A complex number is a number that can be written in the form a + bi, where, a, and b are real numbers, and  “i” is an imaginary number called “iota”. The value of i = (√-1). 

For example, 

1. 2+ 3i is a complex number
2. 3-5i  is a complex number
3. -4+9i is a complex number
4. -2-7i is a complex number

A complex number simply consists of two parts, a real part, and an imaginary part. 

The complex numbers were found by solving the equation x²+1 = 0. The roots of the equation are of form x = ±√-1 and no real roots exist. This is how complex numbers were introduced. 

Find all the solutions of the equation 213-8p³=0 

Solution:

Given that ,

213-8p³ = 0

=>8p³-213 = 0

=>p³-(213/8) = 0        — dividing both sides by 8

=>(p)³-{(213/8)^1/3}³ = 0

Since given equation is cubic, there must be 3 roots .

Let r = p and, s = (213/8)^1/3= 2.986 (approximately)

Now,

r³-s³ = 0

(r-s)(r²+rs+s²) = 0

so, r-s=0 => r=s  [ putting value of s  ]  —eq(1 )

now, (r²+rs+s²)=0

to solve this equation let this is a quadratic equation in r ,  a=1, b=s, c=s²

so, 

d =  b²-4ac = s²-4( 1 )( s² ) = -3s²

r = ( -b ± √d ) / 2a  –formula of roots in quadratic equation

=  ( -s  ± √(-3s²) )/2( 1 )

=  ( -s ± √3si )/2     — where i =   √(-1)

=  s ( -1 ± √3i ) /2     

So , cube root a number having three solution {x³ – s =0} –formula(1)

1.  x = s^1/3
2.  x = s^1/3( -1 + √3i )/2 
3.  x = s^1/3( -1 – √3i )/2

so, solutions for given equation are  s^1/3=(213/8)^1/3=2.986

1.  2.986
2. 2.986( -1 + √3i ) /2
3. 2.986( -1 – √3i ) /2

Polar form of a complex number z = x + i y

z = r (cosθ + i sinθ ) ,

where, r = √( x² + y²) and tanθ = x/y

So , here we have 3 complex numbers 
z₁ = 2.986 + i (0)
z₂ = 2.986 ( -1 + √3i ) /2  
z₃ = 2.986 ( -1 – √3i ) /2  

for z₁, 

r = √( 2.985² + 0²) 

  = 2.985 , 

tanθ = 0/2.985  

=>  θ = 0

z₁ = 2.985( cos(0) + i sin(0) )            –(1)

for z₂ , 

r =  √( (-2.985)² ( 1²+ sqrt(3)² ) ) 

= -2.985 ( 2 ) = -5.97 , 

tanθ = -√(3/1) 

=> θ = -60

z₂ = 5.97( cos(60) – i sin(60) )            –(2)

for z₃, 

r = √( (-2.985)2 ( 1²+ sqrt(3)2 ) ) 

= -2.985 ( 2 ) = -5.97

tanθ =√(3/1 )

=> θ =60

z₃ = 5.97( cos(60) + i sin(60) )            –(3)

Problems based on solution of cubic equation using complex numbers

Problem 1: p³-64=0 , find solutions.

Solution:

from comparing it to x³-s=0 using formula(1) from above explanation
s=64 
s^1/3=64^1/3=4
according to above discussed approach all three solutions are following in the normal form 
x=s^1/3 = 4 
 x=s^1/3( -1 + √3i )/2 = 4( -1 + √3i ) /2 = 2( -1 + √3i ) 
x=s^1/3( -1 – √3i )/2  = 4( -1 – √3i ) /2 = 2( -1 – √3i ) 

Problem 2: p³-343=0,find solutions.

Solution:  

from comparing it to x³-s=0 using formula(1) from above explanation
s=343 
s^1/3=343^1/3=7
according to above discussed approach all three solutions are following in the normal form
x=s^1/3=7
 x=s^1/3( -1 + √3i )/2 = 7( -1 + √3i ) /2
x=s^1/3( -1 – √3i )/2 = 7( -1 – √3i ) /2

Problem 3: p³-27=0,find solutions.

Solution: 

from comparing it to x³-s=0 using formula(1) from above explanation
s=27
s^1/3=27^1/3=3
according to above discussed approach all three solutions are following in the normal form
x=s^1/3=3
 x=s^1/3( -1 + √3i )/2 = 3( -1 + √3i ) /2
x=s^1/3( -1 – √3i )/2 = 3( -1 – √3i ) /2

Problem 4: p³-125=0,find solutions.

Solution:

 from comparing it to x³-s=0 using formula(1) from above explanation
s=125
s^1/3=125^1/3=5
according to above discussed approach all three solutions are following in the normal form
x=s^1/3=5
 x=s^1/3( -1 + √3i )/2 = 5( -1 + √3i ) /2
x=s^1/3( -1 – √3i )/2 = 5( -1 – √3i ) /2