If a + b = \(\frac{3}{4}\) and ab = \(\frac{1}{8}\), then

find a⁴ + b⁴ + ab³ + ba³.

\(\frac{27}{256}\)

a + b = \(\frac{3}{4}\)

Squaring both sides

a² + b² + 2ab = \(\frac{9}{16}\)

a² + b² = \(\frac{9}{16}\) - 2 × \(\frac{1}{8}\)

a² + b² = \(\frac{5}{16}\)

Squaring both sides again in equation

a⁴ + b⁴ = \(\frac{25}{256}\) - 2a²b² 

a⁴ + b⁴ = \(\frac{25}{256}\) - 2 × \(\frac{1}{64}\)

a⁴ + b⁴ = \(\frac{17}{256}\)

⇒ Put the value in below equation

⇒ a⁴ + b⁴ + ab (a² + b²) = \(\frac{17}{256}\) + \(\frac{5}{128}\) = \(\frac{27}{256}\)