Let us factorise the following algebraic expressions:

(x² – 1)² + 8x(x² + 1) + 19x²

As we know that,

a² – b² = (a – b)(a + b)

The given expression can be rewritten as:

(x + 1)²(x – 1)² + 8x(x² + 1) + 19x²

Using (a – b)² = a² + b² – 2ab, and

(a + b)² = a² + b² + 2ab

⇒ (x² + 1 + 2x)(x² + 1 – 2x) + 8x(x² + 1) + 19x²

Let x² + 1 =p

⇒ (p + 2x)(p – 2x) + 8xp + 19x²

⇒ p² – 4x² + 8xp + 19x²

⇒ p² + 8xp + 15x²

⇒ p² + 3xp + 5xp + 15x²

⇒ p(p+3x) + 5x(p + 3x)

⇒ (p + 5x)(p + 3x)

On substituting the value of p, we get,

⇒ (x² + 5x + 1)(x² + 3x + 1)