Power of 3√6 is irrational

Suppose {tex}\sqrt [ 3 ] { 6 }{/tex} be rational number and {tex}\sqrt [ 3 ] { 6 } = \frac { a } { b }{/tex} where {tex}a\ and\ b{/tex} are co-prime and {tex}b\ne0{/tex}
{tex}\Rightarrow ( \sqrt [ 3 ] { 6 } ) ^ { 3 } = \frac { a ^ { 3 } } { b ^ { 3 } }{/tex}

{tex}\Rightarrow 6 = \frac { a ^ { 3 } } { b ^ { 3 } } {/tex}

{tex}\Rightarrow 6 . b ^ { 3 } = a ^ { 3 }{/tex}
{tex}\Rightarrow {/tex} {tex}a^3{/tex} is divisible by {tex}6{/tex} {tex}\Rightarrow{/tex} {tex}a{/tex} is divisible by {tex}6{/tex}.
Let {tex}a = 6c{/tex}
{tex}6b^3 = (6c)^3{/tex}
{tex}\Rightarrow \quad b ^ { 3 } = 36 c ^ { 3 }{/tex}
{tex}\Rightarrow {/tex} {tex}b^3{/tex} is divisible by {tex}6{/tex} {tex}\Rightarrow {/tex} {tex}b{/tex} is  divisible by {tex}6{/tex}.
{tex}\Rightarrow {/tex} {tex}a\ and\ b{/tex} have a common factor i.e, {tex}6{/tex}
{tex}\Rightarrow {/tex} {tex}a\ and\ b{/tex} are not co-prime which is a contradiction
{tex}\therefore \sqrt [ 3 ] { 6 }{/tex} is an irratonal.