If x- y = \(\sqrt {3}\) 6 + xy = \(\frac{6}{xy}\) find x4 + y4

18 19 20 21

21

x + y = \(\sqrt {3}\) Squaring both sides x2 + y2 = \( { \left(\sqrt {3}\right) }^{2} \) + 2xy (using standard identity) Squaring both sides, x4 + y4 = 9 + 4x2y2 + 12xy - 2x2y2  x4 + y4 = 9 + 2x2y2 + 12xy = 9 + 2xy(xy + 6) = 9 + 2xy × \(\frac{6}{xy}\) x4 + y4 = 9 + 12 = 21