# What is the HCF of 8(x^5 - x^3 + x) and 28(x^6 + 1)

LCM HCF
What is the HCF of 8(x^{5} - x^{3} + x) and 28(x^{6} + 1)?

- (x
^{4} + 1 - x^{2})
- 4(x
^{4} + 1 - x^{2})
- 2(x
^{4} + 1 - x^{2})
- None

### Answer

28(x^{6} + 1) = 28(x^{2})^{3} + 1^{3} = 7×4×(x^{2} + 1)(x^{4} + 1 - x^{2})

= 7(x^{2} + 1)×4(x^{4} + 1 - x^{2})

Also, 8 (x^{5} - x^{3} + x) = 2x × 4(x^{4} + 1 - x^{2})

So, HCF = 4(x^{4} + 1 - x^{2})

**The correct option is B.**