[Guest]

Find ax⁵+by⁵ given that:

ax +by = 3

ax²+by²= 7

ax³+by³=16

ax⁴+by⁴=42

(axⁿ+byⁿ)(x+y)=(axⁿ⁺¹+byⁿ⁺¹)+(xy)(ax^n-1+by^n-1)

*this is neither a problem i created nor my homework

[Guest]

Find ax⁵+by⁵ given that:

ax +by = 3

ax²+by²= 7

ax³+by³=16

ax⁴+by⁴=42

(axⁿ+byⁿ)(x+y)=(axⁿ⁺¹+byⁿ⁺¹)+(xy)(ax^n-1+by^n-1)

*this is neither a problem i created nor my homework

from the hint, let n = 2

7*(x+y) = 16+3*x*y

and n = 3

16*(x+y) = 42 + 7*x*y

These two equations have solutions

-7-sqrt(87) and sqrt(87) - 7

where x is one of these and y is the other.

z = ax^5 + bx^5

42*(x+y) = z + x*y*16

z = (x+y)*42 - 16*x*y = (-14)*42 -16*(-7-sqrt(87))*(-7+sqrt(87))

= -14*42 - 16*(49-87) = 20

btw:

a = 5.864681464178368e-05

b = 1.289415037395884

Edited July 17, 2010 by mmiguel1

[Guest]

from the hint, let n = 2

7*(x+y) = 16+3*x*y

and n = 3

16*(x+y) = 42 + 7*x*y

These two equations have solutions

-7-sqrt(87) and sqrt(87) - 7

where x is one of these and y is the other.

z = ax^5 + bx^5

42*(x+y) = z + x*y*16

z = (x+y)*42 - 16*x*y = (-14)*42 -16*(-7-sqrt(87))*(-7+sqrt(87))

= -14*42 - 16*(49-87) = 20

absolutely!

[bonanova]

Solve the equations [n=2] and [n=3] to get

(x+y) = -14 and (xy) = -38.

Plug these values into the equation [n=4] to get

ax⁵ + by⁵ = 20.