given that ax+by=3,ax2+by2=7,ax3+by3=16,ax4+by4=42.Find ax5+by5 value.a,b,x,y are positive integers

Hello student,
Please find the answer to your question below
We have ax³ + by³ = 16,
so (ax³+ by³)(x + y) = 16(x + y)
and thusax⁴+ by⁴ + xy(ax²+ by²) = 16(x + y)
42 + 7xy = 16(x + y)
From ax²+ by² = 7, we have
(ax²+ by²)(x + y) = 7(x + y)
so ax³+ by³ + xy(ax²+ by²) = 7(x + y).
This implies to
16 + 3xy = 7(x + y)
We can now solve for x + y and xy from (1) and (2) to find x + y=-14 and xy=38
Thus we have
(ax⁴+ by⁴)(x + y) = 42(x + y)
and so ax⁵+ by⁵ + xy(ax³ + by³) = 42(x + y).
Finally, it follows that
ax⁵+ by⁵=42(x+y)-16xy=20