If 8a3 + 27b3 = 16 and 2a + 3b= 4, then find the value of 16a4 + 81b4.

8a³ + b³ = 16 and 2a+ b = 4

we know -

a³ + b³ = (a + b)(a² - ab + b²)

(a + b)² = (a² + 2ab + b²)

= (8a³ + b³) = (2a + b)(4a² - 2ab + b²)

= 16 = 4 × (4a² - 2ab + b²)

⇒ 4a² - 2ab + b² = 4

⇒ (2a + b)² = 4a² + 4ab + b²

⇒ 16 = 4a² + 4ab + b²

Solving,

= 6ab = 12

= ab = 2

Then,

= 4a² + b² = 8

Then,

16a⁴ + b⁴ =

 (4a² + b²)² - 8a²b²

8² - 8 × (2)² = 32

16a⁴ + b⁴ = 32