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In given expression

Take out the common factor,

[3 × 3 × 3 × 3 × x × x × x × x - 11 × 11 × x × x]

⇒ x × x[3 × 3 × 3 × 3 × x × x - 11 × 11]

⇒ x²[81x² – 121]

In expression 81x² - 121

Both terms are perfect square

⇒ 81x² = 9x × 9x

⇒ 121 = 11 × 11

∴ 81x² – 121 Seems to be in identity a²-b² = (a + b)(a-b)

Where a = 9x and b = 11;

81x² – 121 = (9x-11)(9x + 11)

Hence the factors of 81x⁴ – 121x² are x²,(9x-11) and (9x + 11)