Question

prove cube root 6 is irrational

Answer 1

Hi guys ☺,
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(this answer helps me to forget you Ayushi)
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Here is your answer,
The answers is in Attachment.
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(ignore my grammatical mistakes...)
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Answer 2

Assume cube root 6 is rational then
let cube root 6 = a/b where ( a and b are co Prime and b not =0)
cubing both sides :6=a^3/b^3
a^3=6b^3
a^3= 2(3b^3)
Therefore 2 divides a^3 or a^2 *a by Euclid division lemma if a prime number divides the product of two integers then it must divide one of the two integers,
since all the terms here are the same
we conclude that 2 divides a
now there are exist an integer k such that
a=2 k
substituting 2 k in the above equation
8k^3=6b^3
b^3=2{(2k^3)/3}
therefore 2 divides b^3. using the same logic as above .
2 divides b .
hence 2 is common factor of both a and b.
but this is a construction of the fact that a and b are co prime numbers.
therefore the initial assumption is wrong
HOPE IT WILL HELP YOU MY FRIEND:)