Question

There are six different sixth roots of 64. That is, there are six complex numbers that solve x^6=64

Please help

Answer 1

There are two possible solutions:

1)
When we know that:


`\sqrt[n]{z}=\sqrt[n]\Big(\cos(\phi+2k\pi)/(n)+i\sin(\phi+2k\pi)/(n)\Big)\qquad\text{for}\quad k=0,1,\ldots,n-1`

when
`z=|z|(\cos\phi+i\sin\phi)`

then:


`x^6=64\quad|\sqrt[6]{(\ldots)}\\\\x=\sqrt[6]{64}`

For
`x=64` we have:


`64=|64|(\cos0+i\sin0)\quad\Rightarrow\quad \phi=0`

and:


`\sqrt[6]{64}=\sqrt[6]{64}\Big(\cos(0+2k\pi)/(6)+i\sin(0+2k\pi)/(6)\Big)\qquad\text{for}\quad k=0,1,\ldots,5\\\\\\ \sqrt[6]{64}=2\Big(\cos(k\pi)/(3)+i\sin(k\pi)/(3)\Big)\qquad\text{for}\quad k=0,1,\ldots,5`

k = 0


`2\Big(\cos(0)/(3)+i\sin(0)/(3)\Big)=2(1+0i)=2\cdot1=\boxed{2}`

k = 1


`2\Big(\cos(\pi)/(3)+i\sin(\pi)/(3)\Big)=2\Big((1)/(2)+i(√(3))/(2)\Big)=\boxed{1+i√(3)}`

k = 2


`2\Big(\cos(2\pi)/(3)+i\sin(2\pi)/(3)\Big)=2\Big(-(1)/(2)+i(√(3))/(2)\Big)=\boxed{-1+i√(3)}`

k = 3


`2\Big(\cos(3\pi)/(3)+i\sin(3\pi)/(3)\Big)=2\Big(\cos\pi+i\sin\pi\Big)=2(-1+0i)=\boxed{-2}`

k = 4


`2\Big(\cos(4\pi)/(3)+i\sin(4\pi)/(3)\Big)=2\Big(-(1)/(2)-i(√(3))/(2)\Big)=\boxed{-1-i√(3)}`

k = 5


`2\Big(\cos(5\pi)/(3)+i\sin(5\pi)/(3)\Big)=2\Big((1)/(2)-i(√(3))/(2)\Big)=\boxed{1-i√(3)}`

So the answer is:


`x=\{2,\,1+i√(3),\,-1+i√(3),\,-2,\,-1-i√(3),\,1-i√(3)\}`

2)
We don't know method (1). If so, we could use following identities:


`(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)`

There will be:


`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)`

Now, we complete the square for:


`x^2-2x+4=x^2-2x+1+3=(x^2-2x+1)+3=(x-1)^2+3=\\\\=(x-1)^2+(√(3))^2=(x-1)^2-(-1)(√(3))^2=(x-1)^2-i^2(√(3))^2=\\\\=(x-1)^2-(i√(3))^2=\text{from (1)}=\boxed{(x-1-i√(3))(x-1+i√(3))}`

and for:


`x^2+2x+4=x^2-2x+1+3=(x^2+2x+1)+3=(x+1)^2+3=\\\\=(x+1)^2+(√(3))^2=(x+1)^2-(-1)(√(3))^2=(x+1)^2-i^2(√(3))^2=\\\\=(x+1)^2-(i√(3))^2=\text{from (1)}=\boxed{(x+1-i√(3))(x+1+i√(3))}`

When we return to
`(\star)`:


`(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=0\\\\(x+2)(x-1-i√(3))(x-1+i√(3))(x-2)(x+1-i√(3))(x+1+i√(3))=\\=0`

And we have answer:


`x=\{-2,\,1+i√(3),\,1-i√(3),\,2,\,-1+i√(3),\,-1-i√(3)\}`