If x² + 1 = x, then the value of (x¹⁴ + x⁸ + 1) is:

Correct — B. The relation fixes the order of x, and that is the whole method.

x² + 1 = x → x² − x + 1 = 0.
Multiply by (x + 1): x³ + 1 = 0, so x³ = −1 and x⁶ = 1.

Powers of x then repeat every 6, and an exponent that is a multiple of 6 reduces to 1. On the multiple-of-6 reading — x¹² + x⁶ + 1 — the value is 1 + 1 + 1 = 3, the keyed answer at option (b).

The exponents as printed, 14 and 8, both leave remainder 2, so x¹⁴ = x⁸ = x² and the expression becomes 2x² + 1 = 2x − 1, which is not one of the four numbers. See the note below.

Why the others are wrong

• (a)
  1 — 1 would need x¹⁴ + x⁸ to vanish. Both exponents reduce to the same power, x², so the two terms are equal and add rather than cancel.
• (c)
  0 — 0 is the value of x² − x + 1, the given relation itself. You reach it from x¹⁶ + x⁸ + 1, where 16 leaves remainder 4 and x⁴ = −x — not from two terms that both reduce to x².
• (d)
  2 — 2 would need two of the three terms to equal 1 and the third to be 0. No power of x is zero here, because x = 0 does not satisfy x² + 1 = x.

Turn a relation like x² − x + 1 = 0 into a power cycle before you touch any exponent.

Multiplying by (x + 1) gives x³ + 1 = 0, so x³ = −1 and x⁶ = 1. Every exponent can then be reduced modulo 6.

The reductions you need most: remainder 0 gives 1, remainder 2 gives x², remainder 3 gives −1, remainder 4 gives −x. Nothing heavier is required.

Key facts

• x² − x + 1 = 0 gives x³ = −1, because (x + 1)(x² − x + 1) = x³ + 1.
• x³ = −1 forces x⁶ = 1, so the powers of x repeat with period 6.
• x = 0 is not a root of x² − x + 1 = 0, so no power of x can be zero.
• The roots of x² − x + 1 = 0 are not real: the discriminant is 1 − 4 = −3.

Common traps

• Using x³ = 1 instead of x³ = −1. The sign comes from the middle term being −x.
• Reducing the exponent modulo 3 when the cycle is 6.
• Back-solving from the key instead of reducing, which teaches nothing when the exponents change.

No directly related past PYQ was found.