Solve this polynomial?

Question

How to solve this polynomial by using manipulation to first make it quadratic?

x^6 + 16x^3 +64 =0

Answer 1

Two possible routes here. First let y = x^3. Then your equation becomes

y^2 + 16y + 64 = 0

one solution is y=-8, thus you can factor (y+8) and get

y^2 + 16y + 64 = (y+8)(y+8) = (y+8)^2 = 0

thus the only solution is y= -8.

Now you know that y = x^3. Since the function f(x) = x^3 is one to one, you conclude that x is the cubic root of y.

This yields x = -2 as your unique solution.

Now a different more fun way.

Note that 16 = 2*2^3 and 64 = 2^6, we can now write the original equation

x^6 + 16x^3 + 64 = x^6 + 2*2^3 x^3 + 2^6 = 0, or

x^6 + 2*2^3 x^3 + (2^3)^2 = 0. We can thus factor this polynomial directly:

x^6 + 2*2^3 x^3 + (2^3)^2 = (x^3+2^3)^2 = 0, which leads to

x^3 = -2^3, or x = -2.

Answer 2

x^6 + 16x^3 +64 =0

(x^3)^2+16x^3+64=0

(x^3+8)^2=0

x^3+8=0

x^3= -8=8e^(iπ)

x=³√(-8)=³√[8e^(iπ)]

x=2e^[i(π+2kπ)/3], where k=0,1,2

x1=2e^(iπ/3)=2[cos(π/3)+isin(π/3)]

x1=1+i√3

x2=2e^(iπ)=2(cosπ+isinπ)

x2= -2

x3=2e^(i5π/3)=2[cos(5π/3)+isin(5π/3)]

x3= -1-i√3

Answer 3

Solving by chunking :

Let y = x^3

Then you get y^2 + 16y + 64 = 0

Factor : (y + 8) (y + 8) = 0

y = -8

But y really equals x^3 so

x^3 = -8

so x = -2.

Good luck to you !