Question

: Find the cube root of $1$ .
(A) $1$
(B) $2$
(C) ${\text{Does not exist}}$
(D) ${\text{None of these}}$

Answer 1

If a number is multiplied by itself three times, then the resulting value is known as the cube of the given value. So, the cube root of a number is defined as, a value which gives the given value, when it is multiplied by itself for three times. So now we use this concept to solve this question.
So, $a \times a \times a = {a^3}$ which implies that, $\sqrt[3]{{{a^3}}} = a$ .

Complete step by step answer:
Here, we need to find the cube root of $1$ .
So, let $x$ be its cube roots, which means ${x^3} = 1$ .
After transposing, we get, ${x^3} - 1 = 0$
$\Rightarrow {x^3} - {1^3} = 0$
And we know the identity, ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
So, from the identity, we get,
${x^3} - {1^3} = (x - 1)({x^2} + x + {1^2}) = 0$
$\Rightarrow (x - 1)({x^2} + x + 1) = 0$
So here, either the term $(x - 1)$ is equal to zero, or the term $({x^2} + x + 1)$ is equal to zero.
$\Rightarrow x - 1 = 0$ or ${x^2} + x + 1 = 0$
$\Rightarrow x = 1$ or $x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4} }}{2}$ (from the quadratic formula)
For a quadratic equation $a{x^2} + bx + c = 0$ the roots are $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow x = 1$ or $x = \dfrac{{ - 1 \pm 3i}}{2}$ where $i = \sqrt { - 1}$ which is a complex number i.e. it has no value. Its value is imaginary, and it can’t be located in a number line.
We are getting one real value, and two complex values.
So, we can conclude that, cube root of $1$ is $1$ .

So, the correct answer is “Option A”.

Note:
We must take all the possible values and calculate the roots of an equation. The complex answers that we got are also the cube roots of one, but those consist of an imaginary part. But in the options, only real numbers are given, so we considered only the real value that we got. To check whether the answer that we got is correct or not, multiply it by itself three times and verify your answer. A quadratic equation can also be solved by completing the square method (as per our convenience).