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Prove that $(6)^\frac{1}{3}$ is not rational.
Given :
The given number is $(6)^\frac{1}{3}$
To do :
We have to prove that $(6)^\frac{1}{3}$ is not rational.
Solution :
$(6)^\frac{1}{3} = \sqrt [3]{6}$
Let us assume $ \sqrt [3]{6}$ is rational.
Hence, it can be written in the form of $\frac{a}{b}$, where a, b are co-prime(no common factors other than 1), and b is not equal to 0.
$ (6)^\frac{1}{3}=\frac{a}{b}$
$6 = (\frac{a}{b})^3$
$6 = \frac{a^3}{b^3}$
$6 b^3 = a^3$
Therefore, $a^3$ is divisible by 6, and a is also divisible by 6.
So, we can write $a = 6 c$ for some integer c.
Substituting $a = 6 c$,
$6 b^3 = (6 c)^3$
$6 b^3 = 6^3 c^3$
$ b^3 = 6^2 c^3$
Which means $b^3$ is divisible by 6, and so b is also divisible by 6.
Therefore, a and b have 6 as a common factor.
But this contradicts that a and b are co-prime(no common factors other than 1).
So, the assumption $(6)^\frac{1}{3}$ is rational is incorrect.
Therefore, $(6)^\frac{1}{3}$ is not a rational number.
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