Hint: Let y = x^(1/3)

Algebra Level 4

How many complex root/s $x$ that satisfy the equation $\sqrt[3]{x^2} + \sqrt[3]{x} + 1 = 0$

4 3 0 1 2

2 solutions

Paul Ryan Longhas
Apr 17, 2015

Let $y = \sqrt[3]{x}$

Hence, $y^2 + y + 1=0$ . So, $y = \frac{-1 \pm i\sqrt{3}}{2}$

So, $\sqrt[3]{x} = \frac{-1 \pm i\sqrt{3}}{2} \Rightarrow x = (\frac{-1 \pm i\sqrt{3}}{2})^3$

Therefore, $x = 1$ . Re substitute $x. 1 + 1 + 1 \neq 0$ . Therefore, no value of x that satisfy the equation.

Nice solution. For the last step I noted that the potential solutions were

$x^{\frac{1}{3}} = e^{i*\frac{2\pi}{3}}, e^{i*\frac{4\pi}{3}} \Longrightarrow x = e^{i*2\pi}, e^{i*4\pi},$

which in both cases yields a value of $1,$ which as you have observed is not in fact a solution of the original equation.

Brian Charlesworth - 6 years, 1 month ago

Vishwak Srinivasan
Apr 19, 2015

As the hint mentions, take $y= \sqrt[3]{x}$

Hence,

$y^2 + y + 1 = 0$

So $y = \omega , \omega^2$

$y^3 = x$

Implies, $x = \omega^3 , \omega^6 = 1 , 1$

No complex root.

Q.E.D

Isnt 1 a complex number i think that it is a complex number as real numbers are a sub set of complex numbers

avn bha - 5 years, 10 months ago

Avn: while the statement you made is true, sadly x=1 does not satisfy the original equation, as Paul noted above.

John Gilling - 5 years, 7 months ago