Verify:
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(i) x³ + y³ = (x + y) (x² + y² – xy)

We know that,

(x + y)³ = x³ + y³ + 3xy (x + y)

⇒  x³ + y³ = (x + y)³ – 3xy (x + y)

[Taking (x + y) common]


⇒  x³ + y³ = (x + y) [(x + y)² – 3xy)     

⇒  x³ + y³ = (x + y) [(x² + y² + 2xy) – 3xy]               ........[(x + y)² = x² + y² + 2 x y]

⇒  x³ + y³ = (x + y) (x² + y² – xy)

(ii) x³ - y³ = (x - y) (x² + y² + xy)

We know that,

(x - y)³ = x³ - y³ - 3xy (x - y)

⇒  x³ - y³ = (x - y)³ + 3xy (x - y)

[Taking (x - y) common]

⇒  x³ - y³ = (x - y) [(x - y)² + 3xy)

⇒ x³ - y³ = (x - y) [(x² + y² - 2xy) + 3xy]            ......[(x - y)² = x² + y² - 2 x y]

= x³ - y³ = (x - y) (x² + y² + xy)