Question

Find all the values of i^(1/3)

Answer 1

first cube both sides ...
x^3 = -1
x^3 + 1 = 0
(x+1) (x^2 - x + 1)= 0
now u will get 1st value of x = -1
and use quadratic formula on (x^2 - x + 1)

i used 1 instead of i for better understanding

Answer 2

Answer:

The value of $i^{1/3}$ are $x=-i$ and $x=\frac{i \pm\sqrt {5}}{2}$.

Step-by-step explanation:

Let suppose $x=i^{1/3}$

⇒ $x^3=i$

⇒ $x^3-i=0$

⇒ $x^3+(-i)=0$ ...... (1)

As we know, $i^2=-1$.

This implies, $i^2\cdot i=-i$

⇒ $i^3=-i$

Substitute the value $i^3$ for $-i$ in the equation (1) as follows:

⇒ $x^3+i^3=0$

Use the identity, $(a^3+b^3)=(a+b)(a^2-ab+b^2)$

⇒ $(x+i)(x^2-xi+i^2)=0$

⇒ $(x+i)(x^2-xi-1)=0$ (Since $i^2=-1$)

⇒ $x=-i$ and $x^2-xi+i^2=0$

⇒ $x=\frac{-b \pm\sqrt {b^2-4ac}}{2a}$

⇒ $x=\frac{-(-i) \pm\sqrt {-i^2-4(1)(i^2)}}{2}$

⇒ $x=\frac{i \pm\sqrt {1+4}}{2}$ (Since $i^2=-1$)

⇒ $x=\frac{i \pm\sqrt {5}}{2}$

Therefore, the value of $i^{1/3}$ are $x=-i$ and $x=\frac{i \pm\sqrt {5}}{2}$.

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