If x- y = \(\sqrt {3}\)
6 + xy = \(\frac{6}{xy}\)
find x4 + y4
Answer & explanation
Correct answer: option 4
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