The term independent of x in the binomial expansion of (1-x+3x⁵)(2x² - x)⁸ is

Given expansion of (1-x+3x⁵)(2x² - x)⁸
For this, we write general term in the expansion of (2x² - x)⁸
Tr+1 = ⁸Cr(2x²)8-r(-x)r = ⁸Cr28-r(-1)rx16-3r
Now, in the product (1-x+3x⁵)(2x² - x)⁸, the term independent of x is
1 × term not containing x in (2x² - x)⁸ - x × term containing x in (2x² - x)⁸ + 3x⁵ × term containing x⁻⁵ in (2x² - x)⁸ (2)
Now, by (1), term without x, 16-3r = 0 ⇒ r = 16/3 which is not possible.
Hence, there is no term independent of x in (2x² - x)⁸
Now, again by (1), term containing x, 16-3r = 1 ⇒ r = 5
So, by (1), T₆ = ⁸C₅2³(-1)⁵x = -448x
Now, again by (1), term containing x⁻⁵, 16-3r = -5 ⇒ r = 7
So, by (1), T₈ = ⁸C₇2¹(-1)⁷x⁻⁵ = -8x⁻⁵
Put all these values in (2), we get
Term independent of x = 1 × 0 - x(-448x) + 3x⁵(-8x⁻⁵) = 448 - 24 = 424