Perfect Cube, Fourth Powers, Fifth Powers, Sixth Powers!

It is clear that $a$ and $d$ must be of the form $x^6$ and $y^3$ , respectively. Hence, $d-a=y^3-x^6=(y-x^2)(y^2+yx^2+x^4)=61$ . Since 61 is a prime number it follows that $y-x^2=1$ and $y^2+yx^2+x^4=61$ . Now, $x=1$ or $x=2$ since $x^4<61$ and $3^4=81>61$ . Since only $x=2,y=5$ satisfies the above equation $a=64$ and $d=125$ which implies that $b=32$ and $c=625$ . Hence, $c-b=593$ .

Since

a^5 = b^6 = b(b^5)

Then

b = (a/b)^5 = n^5 ...........................(1)

where

a/b = n ...............................................(2)

From Eq (1), Eq (2)

a = n^6 ..............................................(3)

Since

c^3 = d^4 = d(d^3)

Then

d = (c/d)^3 = m^3 ........................(4)

Where

c/d = m ............................................(5)

Since

d - a = 61

Then , using Eq's (3) , (4)

m^3 - n^6 = 61

Then

(m - n^2)(m^2 + m.n^2 + n^4) = 61 = (1)(61)

Then

m - n^2 = 1

i.e. n^2 = m - 1 ...................................... ........(6)

, and m^2 + m.n^2 + n^4 = 61................(7)

From Eq (6), Eq (7)

m^2 + m . (m - 1)+ (m - 1)^2 = 61

Then

m = 5

From Eq (6)

n = 2

From Eq's (1) , (3) , (4) , (5)

b = 32 , a = 64 , d = 125, c = 625

Then

c - b = 625 - 32 = 593