Answer
We will use prove by contradiction technique in order to prove that $\sqrt[3] 2$ is irrational we try to prove that $\sqrt[3] 2$ is rational.
Work Step by Step
here is the step by step solution to this question
Let us assume $\sqrt[3]{2}$ is rational, i.e
$\sqrt[3]{2}=a / b$ (a and b are integers)
Now, by taking cube on both sides
$2=\frac{a^{3}}{b^{3}}$
$ \Rightarrow a^{3}=2 b^{3}$
Now,
$a^3$
is an even integer since
$b^3$
is multiplied by 2 and hence a will also be an even integer. So we can write
$a = 2c$ (let c be another arbitrary constant)Now, cube on both sides.
$a^{3}=8 c^{3}$
$2 b^{3}=a^{3}=8 c^{3}$
$8^{3}=4 c^{3}$
Now, $b^3$ is an even integer since $c^3$ is multiplied by even integer and hence b will also be an even integer.
So we can write b=2d (let d be another arbitrary constant) and this process will go to infinity.
Now since a and b both are even integers, it cannot be reduced as a fraction so what we assume is incorrect and hence $\sqrt[3]{2}$ is irrational.
