An interesting one

ans=20
using the identity - (x+y)(axⁿ + byⁿ) - xy(ax^n-1 + by^n-1) = axⁿ⁺¹ + yⁿ⁺¹

ax+by =3
=>ax²+bxy=3x

ax²+by²=7

from these two:

by(x-y)=3x-7......................................................(1)

from the rest of the equations:

by²(x-y)=7x-16....................................................(2)
and ,

by³(x-y)=16x-42...................................................(3)

or, y=7x-163x-7=16x-427x-16.

or, x²+14x-38=0...................................................(4)

Now, we do not need to solve this equation...

Notice that whatever we had done in case if x could also have been done with y
i.e.
we can get equations:

ax(y-x)=3y-7......................................................(1')

ax²(y-x)=7y-16....................................................(2')
and ,

ax³(y-x)=16y-42...................................................(3')

or, x=7y-163y-7=16y-427y-16.

or, y²+14y-38=0...................................................(4')

So, x and y are roots of the equation t²+14t-38=0

x+y=-14
xy =-38

Now utilizing the fact that

(x+y)(ax⁴+by⁴)=ax⁵+by⁵+xy(ax³+by³)

ax⁵+by⁵=38*16 - 14*42= 20