Question

Prove that 6- root7is irrational
by process​

Answer 1

Answer:

a^6 + a^3 - 8a^3 - 8

= a^3 ( a^3 + 1) - 8 ( a^3 + 1)

= (a^3 - 8)(a^3 + 1)

= [ (a)^3 - (2)^3] [ (a)^3 +(1)^3]

= (a-2)(a^2 + 2a + 4)(a+1)(a^2 - a + 1)

Step-by-step explanation:

Answer 2

Answer:

let us assume that 6-√7as rational

let 6-√7=a/b(a,b are co-primes

-√7=a-6b/b

since a,b,6are integers and a-6b/b is in the form of p/q . so, a-6b/b is a rational

hence,√7 is rational

but the method of contradiction that √7is irrational

so,we conclude that 6-√7 is irrational

so please mark me as brainlist