There are two possible solutions:
1)
When we know that:
[tex]\sqrt[n]{z}=\sqrt[n]{|z|}\Big(\cos\frac{\phi+2k\pi}{n}+i\sin\frac{\phi+2k\pi}{n}\Big)\qquad\text{for}\quad k=0,1,\ldots,n-1[/tex]
when [tex]z=|z|(\cos\phi+i\sin\phi)[/tex]
then:
[tex]x^6=64\quad|\sqrt[6]{(\ldots)}\\\\x=\sqrt[6]{64}[/tex]
For [tex]x=64[/tex] we have:
[tex]64=|64|(\cos0+i\sin0)\quad\Rightarrow\quad \phi=0[/tex]
and:
[tex]\sqrt[6]{64}=\sqrt[6]{64}\Big(\cos\frac{0+2k\pi}{6}+i\sin\frac{0+2k\pi}{6}\Big)\qquad\text{for}\quad k=0,1,\ldots,5\\\\\\
\sqrt[6]{64}=2\Big(\cos\frac{k\pi}{3}+i\sin\frac{k\pi}{3}\Big)\qquad\text{for}\quad k=0,1,\ldots,5
[/tex]
k = 0
[tex]2\Big(\cos\frac{0}{3}+i\sin\frac{0}{3}\Big)=2(1+0i)=2\cdot1=\boxed{2}[/tex]
k = 1
[tex]2\Big(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\Big)=2\Big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Big)=\boxed{1+i\sqrt{3}}[/tex]
k = 2
[tex]2\Big(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\Big)=2\Big(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\Big)=\boxed{-1+i\sqrt{3}}[/tex]
k = 3
[tex]2\Big(\cos\frac{3\pi}{3}+i\sin\frac{3\pi}{3}\Big)=2\Big(\cos\pi+i\sin\pi\Big)=2(-1+0i)=\boxed{-2}[/tex]
k = 4
[tex]2\Big(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\Big)=2\Big(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\Big)=\boxed{-1-i\sqrt{3}}[/tex]
k = 5
[tex]2\Big(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}\Big)=2\Big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\Big)=\boxed{1-i\sqrt{3}}[/tex]
So the answer is:
[tex]x=\{2,\,1+i\sqrt{3},\,-1+i\sqrt{3},\,-2,\,-1-i\sqrt{3},\,1-i\sqrt{3}\}[/tex]
2)
We don't know method (1). If so, we could use following identities:
[tex](1)\quad a^2-b^2=(a+b)(a-b)\\\\(2)\quad a^3-b^3=(a-b)(a^2+ab+b^2)\\\\(3)\quad a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]
There will be:
[tex]x^6=64\\\\x^6-64=0\\\\(x^3)^2-8^2=0 \qquad\text{from (1)}\\\\(x^3+8)(x^3-8)=0\\\\(x^3+2^3)(x^3-2^3)=0\qquad\text{from (2) and (3)}\\\\
(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0\qquad(\star)[/tex]
Now, we complete the square for:
[tex]x^2-2x+4=x^2-2x+1+3=(x^2-2x+1)+3=(x-1)^2+3=\\\\=(x-1)^2+(\sqrt{3})^2=(x-1)^2-(-1)(\sqrt{3})^2=(x-1)^2-i^2(\sqrt{3})^2=\\\\=(x-1)^2-(i\sqrt{3})^2=\text{from (1)}=\boxed{(x-1-i\sqrt{3})(x-1+i\sqrt{3})}[/tex]
and for:
[tex]x^2+2x+4=x^2-2x+1+3=(x^2+2x+1)+3=(x+1)^2+3=\\\\=(x+1)^2+(\sqrt{3})^2=(x+1)^2-(-1)(\sqrt{3})^2=(x+1)^2-i^2(\sqrt{3})^2=\\\\=(x+1)^2-(i\sqrt{3})^2=\text{from (1)}=\boxed{(x+1-i\sqrt{3})(x+1+i\sqrt{3})}[/tex]
When we return to [tex](\star)[/tex]:
[tex](x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0\\\\(x+2)(x-1-i\sqrt{3})(x-1+i\sqrt{3})(x-2)(x+1-i\sqrt{3})(x+1+i\sqrt{3})=\\=0[/tex]
And we have answer:
[tex]x=\{-2,\,1+i\sqrt{3},\,1-i\sqrt{3},\,2,\,-1+i\sqrt{3},\,-1-i\sqrt{3}\}
[/tex]
