Find the vale 0f \(2^8+1\) and \(2^{18} +1\)

Can you now finish the problem ..........

Find the lest and largest cubes betwwen \(2^8+1\) and \(2^{18} +1\)

can you finish the problem........

The value of \(2^8+1=257\)

Now less than 257 the perfect cube is \(6^3\)=216 and \(7^3=343\),which is greater than 257

Now \(2^{18}=(2^6)^3=(64)^3\) which will be the largrst cube less than \(2^{18 }+1\)

Hece the required number of cubes is 64-7+1=58