Explanation:
B We have,
\(x^3+y^3=65\)
\(y^3=65-x^3\)
Now
\(y=f(x)=65-x^3\)
\(y=f(x)=y^3\)
When,

Now,
\(x=0\)
\(y=65\)
Now,
\(y=y^3\)
\(y^3-y=0\)
\(y\left(y^2-1\right)=0\)
\(y(y-1)(y+1)=0\)
\(y=0, y=-1, y=1\)
\(y^3=0, y^3=-1, y^3=1\)
\(65-x^3=0\)
\(x^3=65\)
\(x \neq z\)
Whene \(y=-1\)
\(65-x^3=-1\)
\(x^3=66\)
\(x \neq z\)
When, \(\mathrm{y}=1\)
\(65-x^3=1\)
\(\mathrm{x}=4\) then \(\mathrm{y}=1\)
When \(y=4\) then \(x=1\)
Then \((4,1)\) and \((1,4)\) two order pair are possible.