Question

If x²+X+1=0 then what is the value of x²⁰¹⁵ + x²⁰¹³ + x​

Answer 1

Since
`x^2 + x + 1 = 0` means
`x^2+x = -1`, we have for any
`n` the recursive relation

`x^n (x^2 + x) = x^(n+2) + x^(n+1) = -x^n \implies x^(n+2) = -x^(n+1) - x^n`

Let
`f(x)=x^(2015)+x^(2013)+x`. Substitution using the recursive rule generates an alternating power-reduction pattern:

`f(x) = \left(-x^(2014) - x^(2013)\right) + x^(2013) + x = -x^(2014) + x`

`f(x) = -\left(-x^(2013) - x^(2012)\right) + x = x^(2013) + x^(2012) + x`

`f(x) = \left(-x^(2012) - x^(2011)\right) + x^(2012) + x = -x^(2011) + x`

`f(x) = -\left(-x^(2010) - x^(2009)\right) + x = x^(2010) + x^(2009) + x`

`f(x) = \left(-x^(2009) - x^(2008)\right) + x^(2009) + x = -x^(2008) + x`

`f(x) = -\left(-x^(2007) - x^(2006)\right) + x = x^(2007) + x^(2006) + x`

and so on.

Notice that in the equivalent forms of
`f(x)` involving 3 terms, the largest power of
`x` is a multiple of 3, and the next largest power is 1 less. This means after so many iterations of substitutions, we would end up with

`f(x) = x^3 + x^2 + x`

Then by factorizing, the expression of interest reduces to

`f(x) = x (x^2 + x + 1) = x * 0 = \boxed{0}`