Question

The solutions of :
√a - 1/b - 2c + 3d = 1
a + 1/b² - 4c² - 9d² = 3
a√a - 1/b³ - 8c³ + 27d³ = -5
a² + 1/b⁴ -16c⁴ - 81z⁴ = 15
is ...?

Answer 1

\[\sqrt{a}-\frac{1}{b}-2c+3d=1\]\[a-\frac{1}{b^2}-4c^2-9d^2=3\]\[a\sqrt{a}-\frac{1}{b^3}-8c^3+27d^3=-5\]\[a^2+\frac{1}{b^4}-16c^4-81d^4=15\] ?

Answer 2

let \[A=\sqrt{a}\]\[B=\frac{1}{b}\]\[C=2c\]\[D=3d\]and see here
http://openstudy.com/study#/updates/5025344ce4b0d38989b7378f

Answer 3

@tanjung

Answer 4

waw, very spectaculer...thanku very much mukushla

Answer 5

yw :) this is a nice problem