​If $$a^2+1=a$$​, then the value of $$a^12+a^6+1$$​ is

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Correct option is D

1. Given Equation:

​$$a^2 + 1 = a$$​

Rearranging terms:

​$$a^2 - a + 1 = 0$$​

2. Key Property:

From a^2 - a + 1 = 0 , multiplying through by a :

​$$a^3 - a^2 + a = 0$$​

Using a^2 = a - 1 from the given equation:

​$$a^3 = (a - 1)a + a = a^2 - a + 1 = 0$$​

Hence:

​$$a^3 = 1$$​

3. Simplifying $$a^{12} + a^6 + 1 :$$​​

Using $$a^3$$​ = 1 , the higher powers of a cycle as follows:

​$$a^4 = a, \quad a^5 = a^2, \quad a^6 = (a^3)^2 = 1, \text{ and so on.}$$​

Therefore:

​$$a^{12} = (a^3)^4 = 1, \quad a^6 = 1$$​

Substituting these values:

​$$a^{12} + a^6 + 1 = 1 + 1 + 1 = 3$$​

Final Answer:

​$$\boxed{3}$$​