If `x+1/x=3`, then the value of `x^6+1/x^6`is.

Explanation

`(x+1/x)^2=x^2+1/x^2+2xx x xx1/x`
`=> (3)^2=x^2+1/x^2+2`
`=> 9=x^2+1/x^2+2`
`=>x^2+1/x^2+2=9`
`=>x^2+1/x^2=9−2=7`
`∴ x^2+1/x^2=7`
`[x^2 +(1/x^2)]^3=7^3`
LHS is in the form of `(a+b)^3=a^3+b^3+3ab(a+b)`
Hence, `[x^2]^3+ [(1/x^2)]^3+3(x^2)xx(1/x^2)[x^2+(1/x^2)]=343`
`=> x^6+(1/x^6)+3×7=343`
`=> x^6+(1/x^6)+21=343`
` x^6+(1/x^6)=343−21=322`