Question

Find all the roots for the given equation.
$\sf \implies {x}^{6} = - 1$​

Answer 1

**Edit: Two roots were missing, and $\sf{S_{60}}$ was calculated again.

Question: Find all roots.

 An equation of 3rd or higher power requires the zero product rule.

 To apply the zero product, one hand should equal to zero.

$\implies \sf{x^6+1=0}$

 

 Now let's find the factorization, based on identities.

$\implies \sf{(x^2)^3+1=0}$

$\implies \sf{(x^2+1)(x^4-x^2+1)=0}$

 

 It is either:-

• $\sf{x^2+1=0}$ or
• $\sf{x^4-x^2+1=0}$.

 

 In the first equation, both roots are $\sf{x=\pm i}$.

 The second equation is irreducible over rational numbers. Let's factorize it over reals.

$\implies \sf{(x^2+1)^2-(\sqrt{3} x)^2=0}$

$\implies \sf{(x^2+\sqrt{3} x+1)(x^2-\sqrt{3} x+1)=0}$

 For the first factor $\sf{x=\dfrac{-\sqrt{3} \pm i }{2}}$

 For the second factor $\sf{x=\dfrac{\sqrt{3} \pm i }{2} }$

 

 There are a total of 6 complex roots. Hence all roots were found.

 $\sf{x=\pm i}$ or $\sf{x=\dfrac{-\sqrt{3} \pm i}{2} }$ or $\sf{x=\dfrac{\sqrt{3} \pm i}{2} }$.

Learn more:

 How to find $\sf{\omega+2\omega^2+3\omega^3+...+n\omega^n}$ where $\sf{\omega}$ are any roots of $\sf{\omega^6=-1}$.

 Let the sum of nth terms be $\sf{S_n=\sum^{n}_{k=1}k\omega^k}$.

• $\sf{S_n=\omega+2\omega^2+3\omega^3+...+n\omega^n}$
• $\sf{\omega S_n=0+\omega^2+2\omega^3+...+(n-1)\omega^n+n\omega^{n+1}}$

 After subtracting each other, the first nth terms are a G.P.

$\implies \sf{(1-\omega)S_n=(\omega+\omega^2+\omega^3+...+\omega^n)-n\omega^{n+1}}$

$\implies \sf{(1-\omega)S_n=\dfrac{\omega(1-\omega^n)}{1-\omega} -n\omega^{n+1}}$

$\implies \sf{(1-\omega)S_n=\dfrac{\omega-\omega^{n+1}-n\omega^{n+1}+n\omega^{n+2}}{1-\omega}}$

$\implies \sf{S_n=\dfrac{\omega-\omega^{n+1}-n\omega^{n+1}+n\omega^{n+2}}{(1-\omega)^2}}$

$\therefore \sf{S_n=\dfrac{\omega\{n\omega^{n+1}-(n+1)\omega^n+1\}}{(1-\omega)^2} }$

Let's find the sum of 60 terms of the series, using $\sf{\omega^6=-1}$.

$\sf{S_{60}=\dfrac{\omega\{60\omega^{61}-61\omega^{60}+1\}}{(1-\omega)^2} }$

$\implies \sf{S_{60}=\dfrac{60\omega^{62}-61\omega^{61}+\omega}{(1-\omega)^2}}$

$\implies \sf{S_{60}=\dfrac{60\omega^2-61\omega+\omega}{\omega^2-2\omega+1} }$

$\implies \sf{S_{60}=\dfrac{60\omega^2-60\omega}{\omega^2-2\omega+1} }$

$\implies \sf{S_{60}=60(1+\dfrac{\omega-1}{\omega^2-2\omega+1}) }$

$\implies \sf{S_{60}=60(1+\dfrac{1}{\omega-1} )}$

$\implies \sf{S_{60}=60\{1+\dfrac{60(\omega^2+\omega+1)(\omega^3+1)}{(\omega-1)(\omega^2+\omega+1)(\omega^3+1)} \}}$

$\implies \sf{S_{60}=60(1+\dfrac{\omega^5+\omega^4+\omega^3+\omega^2+\omega+1}{\omega^6-1} )}$

$\implies \sf{S_{60}=60(1+\dfrac{\omega^5+\omega^4+\omega^3+\omega^2+\omega+1}{-2} )}$

$\therefore \sf{S_{60}=-30 (\omega^5+\omega^4+\omega^3+\omega^2+\omega-1)}$

Answer 2

$\sf{x^6=-1}$

Rewrite the Equation using u = x² and u³ = (x²)³ = x⁶

$\to\sf{u^3=-1}$

Now solve the Equation u³ = -1

• For $\sf{g^{3}(x)=f(a)}$ the solutions are $\sf{\displaystyle g(x)=\sqrt[\sf{3}]{\sf{f(a)}}, \sqrt[\sf{3}]{\sf{f(a)}} \frac{-1-\sqrt{3} i}{2}, \sqrt[\sf{3}]{\sf{f(a)}} \frac{-1+\sqrt{3} i}{2}}$

$\to\sf{\displaystyle u=\sqrt[\sf{3}]{\sf{-1}},\:u=\sqrt[\sf{3}]{\sf{-1}}\frac{-1+\sqrt{3}i}{2},\:u=\sqrt[\sf{3}]{\sf{-1}}\frac{-1-\sqrt{3}i}{2}}$

Substitute back u = x² and solve for x

$\rule{315}{1}$

Solve $\sf{x^2=\sqrt[\sf{3}]{\sf{-1}}}$

• We know $\sf{\sqrt[\sf{n}]{\sf{-1}}=-1,\:\sf{if\:}n\sf{\:is\:odd}}$

$\to\sf{x^2=-1}$

• For $\sf{x^2 = f(a)}$ the solutions are $\sf{x=\sqrt{\sf{f(a)}},-\sqrt{\sf{f(a)}}}$

$\boxed{\to\sf{x=\sqrt{-1},\:x=-\sqrt{-1}}}$

$\rule{315}{1}$

Solve $\sf{\displaystyle x^2=\sqrt[\sf{3}]{\sf{-1}}\frac{-1+\sqrt{3}i}{2}}$

• Substitute x = a + bi

$\to\sf{\displaystyle\left(a+bi\right)^2=\sqrt[\sf{3}]{\sf{-1}}\frac{-1+\sqrt{3}i}{2}}$

$\to\sf{\displaystyle\left(a^2-b^2\right)+2iab=-\frac{\sqrt[\sf{3}]{\sf{-1}}}{2}+i\frac{\sqrt[\sf{3}]{\sf{-1}}\sqrt{3}}{2}}$

• Complex numbers can be equal only if their real and imaginary parts are equal. Rewrite as system of equations :

$\to\begin{bmatrix}\sf{a^2-b^2=-\dfrac{\sqrt[\sf{3}]{\sf{-1}}}{2}}\\\\ \sf{2ab=\dfrac{\sqrt[\sf{3}]{\sf{-1}}\sqrt{3}}{2}}\end{bmatrix}$

• On Solving it further, We get

$\to\begin{pmatrix}\sf{a=-\dfrac{\sqrt{3}}{2}},\:&\sf{b=\dfrac{1}{2}}\\\\ \sf{a=\dfrac{\sqrt{3}}{2}},\:&\sf{b=-\dfrac{1}{2}}\end{pmatrix}$

• Substitute back x = a + bi

$\boxed{\to\sf{\displaystyle x=-\frac{\sqrt{3}}{2}+\frac{1}{2}i,\:x=\frac{\sqrt{3}}{2}-\frac{1}{2}i}}$

$\rule{315}{1}$

Solve $\sf{\displaystyle x^2=\sqrt[\sf{3}]{\sf{-1}}\frac{-1-\sqrt{3}i}{2}}$

• Substitute x = a + bi

$\to\sf{\displaystyle\left(a+bi\right)^2=\sqrt[\sf{3}]{\sf{-1}}\frac{-1-\sqrt{3}i}{2}}$

$\to\sf{\displaystyle\left(a^2-b^2\right)+2iab=-\frac{\sqrt[\sf{3}]{\sf{-1}}}{2}-i\frac{\sqrt[\sf{3}]{\sf{-1}}\sqrt{3}}{2}}$

• Complex numbers can be equal only if their real and imaginary parts are equal. Rewrite as system of equations :

$\to\begin{bmatrix}\sf{a^2-b^2=-\dfrac{\sqrt[\sf{3}]{\sf{-1}}}{2}}\\\\ \sf{2ab=-\dfrac{\sqrt[\sf{3}]{\sf{-1}}\sqrt{3}}{2}}\end{bmatrix}$

• On Solving it further, We get

$\to\begin{pmatrix}\sf{a=\dfrac{\sqrt{3}}{2}},\:&\sf{b=\dfrac{1}{2}}\\\\ \sf{a=-\dfrac{\sqrt{3}}{2}},\:&\sf{b=-\dfrac{1}{2}}\end{pmatrix}$

• Substitute back x = a + bi

$\boxed{\to\sf{\displaystyle x=\frac{\sqrt{3}}{2}+\frac{1}{2}i,\:x=-\frac{\sqrt{3}}{2}-\frac{1}{2}i}}$

$\rule{315}{1}$

The Final Solutions Are,

$\boxed{\to\sf{\displaystyle x=i,\:x=-i,\:x=-\frac{\sqrt{3}}{2}+\frac{1}{2}i,\:x=\frac{\sqrt{3}}{2}-\frac{1}{2}i,\:x=\frac{\sqrt{3}}{2}+\frac{1}{2}i,\:x=-\frac{\sqrt{3}}{2}-\frac{1}{2}i}}$