The HCF of (x⁴ − y⁴), (x⁸ − y⁸) and (x² − y²) is:

Question

Question

This question was previously asked in

RRB Group D 8 Sept 2022 Shift 3 Official Paper

Answer 1

Given:

(x4 − y4), (x8 − y8) and (x2 − y2)

Concept Used:

HCF is the highest common factor in each of the terms to divide it completely.

a² – b² = (a + b)(a – b)

Calculation:

Factorizing each of the terms,

⇒ x8 − y⁸

= (x⁴ + y⁴)(x⁴ – y⁴)

= (x4 + y4)(x² + y²)(x² – y²)

⇒ x8 − y8 = (x4 + y4)(x2 + y2)(x + y)(x – y)

For the second term,

⇒ x4 − y⁴

= (x2 + y2)(x2 – y2)

⇒ x4 − y4 = (x2 + y2)(x + y)(x – y)

For the third term,

⇒ x2 − y² = (x + y)(x – y)

In all the prime factorizations, the common factor is (x + y)(x – y).

Therefore, the HCF of (x4 − y4), (x8 − y8) and (x2 − y2) is (x + y)(x – y).