Factor the difference of two squares: 81x² − 121y².

Step-by-step solution

1. 1

   Confirm the pattern applies. There are exactly two terms, they are subtracted, and each is a perfect square: $81x^2 = (9x)^2$ since $9^2 = 81$, and $121y^2 = (11y)^2$ since $11^2 = 121$.

2. 2

   Name a and b. With $a = 9x$ and $b = 11y$ the expression is exactly $a^2 - b^2$.

3. 3

   Apply the identity. $a^2 - b^2 = (a - b)(a + b)$, so the factorisation is $(9x - 11y)(9x + 11y)$.

4. 4

   Check that nothing more comes out. Each factor is linear in two variables with coprime coefficients, so neither can be broken down further.

5. 5

   Verify by expanding. $(9x - 11y)(9x + 11y) = 81x^2 + 99xy - 99xy - 121y^2 = 81x^2 - 121y^2$; the cross terms cancel, as they always do for conjugates.