If x^2 + 1/x^2 = 1 for a non zero real number x, then what is the value of the expression x^48 + x^42 + x^36 + x^30 + x^24 + x^18 + x^12 + x^6 + 1?

Introduction / Context:
This algebraic powers question tests familiarity with cyclic patterns of powers when a base satisfies a special relation such as x^2 + 1/x^2 = 1. Instead of computing each term separately, we exploit the structure of the equation to deduce repeating behavior for even powers of x and then sum them efficiently. Such problems often appear in higher level algebra and aptitude tests to check deeper understanding of exponents and identities.

Given Data / Assumptions:

x is a non zero real number (or possibly complex), and x^2 + 1/x^2 = 1.
We need to evaluate S = x^48 + x^42 + x^36 + x^30 + x^24 + x^18 + x^12 + x^6 + 1.

Concept / Approach:
Let t = x^2. Then t + 1/t = 1. This implies t^2 − t + 1 = 0, a quadratic equation whose roots are complex numbers on the unit circle. Powers of t will repeat with a certain period because t is a root of unity up to sign. We want powers of x that are multiples of 6, 12, 18, and so on, which correspond to integer powers of t. Observing the pattern in t^n for different n allows us to compute each term in the sum S as either 1 or −1 and then add them.

Step-by-Step Solution:
Let t = x^2. Then x is non zero, so t is non zero.Given x^2 + 1/x^2 = 1 becomes t + 1/t = 1.Multiply by t: t^2 + 1 = t, so t^2 − t + 1 = 0.From this quadratic, t is a complex number of modulus 1, and it satisfies t^2 = t − 1.One can show that t^6 = 1, so powers of t repeat every 6 steps.Now rewrite each term in S in terms of t: x^48 = (x^2)^24 = t^24, x^42 = (x^2)^21 = t^21, and so on.Compute exponents modulo 6: 24 mod 6 = 0, 21 mod 6 = 3, 18 mod 6 = 0, 15 mod 6 = 3, 12 mod 6 = 0, 9 mod 6 = 3, 6 mod 6 = 0, 3 mod 6 = 3, and 0 mod 6 = 0 for the constant 1 term, considered as t^0.For t satisfying t^6 = 1 and t^2 − t + 1 = 0, we have t^0 = 1 and t^3 = −1.Thus terms with exponent multiple of 6 contribute 1, and terms with exponent congruent to 3 mod 6 contribute −1.In S, x^48, x^36, x^24, x^12, and 1 correspond to t^0 and give 1 each.Terms x^42, x^30, x^18, and x^6 correspond to t^3 and give −1 each.So S = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + 1 = 1.

Verification / Alternative check:
We can check with a specific complex solution of t^2 − t + 1 = 0, such as t = (1 + i√3)/2, which has modulus 1 and satisfies t^6 = 1. Evaluating t^0 and t^3 for this t confirms that t^0 = 1 and t^3 = −1. Substituting into the pattern for S yields exactly the same alternating sum of ones and minus ones equal to 1, confirming the reasoning.

Why Other Options Are Wrong:
The values −9, 0, 9, and 3 would correspond to different counts of 1 and −1 terms in the sum or to an incorrect assumption that all terms are the same. Our careful modular exponent analysis shows exactly five terms equal to 1 and four terms equal to −1, giving 1. Any other total would contradict the algebraic structure implied by t^2 − t + 1 = 0.

Common Pitfalls:
Students may try to compute large powers directly or forget to switch to t = x^2. Another frequent error is to assume that because x^2 + 1/x^2 = 1, x itself must be 1, which is not true here since x can be complex. The correct path is to exploit the quadratic relation for t and observe the periodic behavior of t^n.

Final Answer:
The value of x^48 + x^42 + x^36 + x^30 + x^24 + x^18 + x^12 + x^6 + 1 is 1.