If (x – 1) and (x + 1) are two factors of the polynomial x 6 – 1, then its other factors are

Treat x⁶ − 1 as a difference of squares first: (x³)² − 1 = (x³ − 1)(x³ + 1). Each cube splits further. x³ − 1 = (x − 1)(x² + x + 1) and x³ + 1 = (x + 1)(x² − x + 1). So x⁶ − 1 = (x − 1)(x + 1)(x² + x + 1)(x² − x + 1). The two given factors, (x − 1) and (x + 1), are accounted for, and the remaining pair is (x² − x + 1) and (x² + x + 1). That is Option C.

Why the others fail:

• A and B: x² + 1 does not divide x⁶ − 1. Put x = i: i⁶ − 1 = −2, not 0.
• D: The roots of x⁶ − 1 are the sixth roots of unity, all with modulus 1. The roots of x² − x − 1 are (1 ± √5)/2, which are not of modulus 1, so this quadratic cannot be a factor.

Core concept: factorising xⁿ − 1 using difference of squares and sum/difference of cubes. This is basic algebra for CSAT and the quantitative section, and the roots-of-unity view is useful for later work on polynomials.

Trap: Aspirants pick A or B by reasoning from x⁴ − 1 = (x² − 1)(x² + 1). That identity is for x⁴ − 1, not x⁶ − 1.