2 Find the roots of the following function fx x6 16x3 64

2).

Find the roots of the following function:

f(x) = x⁶ + 16x³ + 64 = 0

Solution

Given that

f(x) = x⁶ + 16x³ + 64 = 0

Let u = x³ , u² = x⁶

Substitute u ,u² in given equation

    x⁶ + 16x³ + 64 = 0

u² + 16u + 64 = 0

( u+8) (u+8) = 0

u = -8

Substitute back u = x³

x³ = -8

We know that

For x³ = f(a) the solutions are, x =  ³f(a) ,  ³f(a).( (-1 - 3i)/2 ) , ³f(a).( (-1 + 3i)/2 )

  x³ = - 8

Here, f(a) = - 8   

  x =  ³-8 ,   ³-8.( (-1 - 3i)/2 ) ,   ³-8.( (-1 + 3i)/2 )

x = - 2 , -2 . ( (-1 - 3i)/2 ) , -2 . ( (-1 + 3i)/2 ) [ since, ³-8 = -³(2)³ = - (2³)^1/3 = - 2 ]

x = - 2 , - ( -1 - 3i) , - ( -1 + 3i)

x = - 2 , 1 + 3i , 1 - 3i

Therefore,

The roots are x =  - 2 , 1 + 3i , 1 - 3i