Question

y3/x.y/x both are under root​

Answer 1

Answer:

Put x+y=:u, x−y=:v; then u and v are integers of the same parity. In the new variables the equation becomes

1

8

(6u2v+2v3)=

1

4

(u2−v2)+61 ,

which can be written as

27u2+(3v+2)2=

6584

3v−1

.

It follows that v>1 and that 3v−1 must divide 6584=8⋅823; whence v∈{1,3,549,2195}. 549 and 2195 are obviously too large, and v:=3 leeds to u2=26 which has no integer solutions. It remains v=1 which leads to u2=121 or u=±11. Therefore we only have the two solutions (x1,y1):=(6,5) and (x2,y2):=(−5,−

Step-by-step explanation:

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