If x + y = \(\frac{3}{4}\) and xy = \(\frac{1}{8}\), then the value of x⁴ + y⁴ + xy³ + yx³ is?

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

x + y = \(\frac{3}{4}\)

on squaring both sides → x² + y² + 2xy = \(\frac{9}{16}\)

x² + y² = \(\frac{9}{16}\) - \(\frac{2}{8}\)

x² + y² = \(\frac{5}{16}\) 

squaring both sides again

x⁴ + y⁴ + 2x²y² = \(\frac{25}{256}\)

x⁴ + y⁴ = \(\frac{25}{256}\) - \(\frac{1}{32}\)

x⁴ + y⁴ = \(\frac{17}{256}\)

Now these values in the eq.:

⇒ x⁴ + y⁴ + xy³ + yx³

= x⁴ + y⁴ + xy (x² + y²)

= \(\frac{17}{256}\) + \(\frac{1}{8}\) (\(\frac{5}{16}\)) = \(\frac{17}{256}\) + (\(\frac{5}{128}\)) = \(\frac{27}{256}\)