Let a - d, a, a + d, a + 2d be four terms of an arithmetic progression with integer entries and a, a + d, a + 2d, a + 3d be another four terms of the same arithmetic progression. Let $$x = a(a - d) (a + d) (a + 2d) + d^4$$ and $$y = a(a + d) (a + 2d) (a + 3d) + d^4$$. Then x + y is equal to ?

Given that $$x = a(a - d) (a + d) (a + 2d) + d^4$$ 

$$x=a\left(a-d\right)\left(a+d\right)\left(a+2d\right)+d^4$$

$$=a\left(a+d\right)\left(a-d\right)\left(a+2d\right)+d^4$$

$$=\left(a^2+ad\right)\left(a^2+2ad-ad-2d^2\right)+d^4$$

Let $$t=a^2+ad$$

$$x=\left(t\right)\left(t-2d^2\right)+d^4$$

$$x=t^2-2td^2+d^4$$

$$x=\left(t-d^2\right)^2$$

Substituting $$t=a^2+ad$$

$$x=\left(a^2+ad-d^2\right)^2$$.........(1)

Given that,  $$y = a(a + d) (a + 2d) (a + 3d) + d^4$$

$$y=a\left(a+d\right)\ \left(a+2d\right)\ \left(a+3d\right)+d^4$$

$$=a\left(a+3d\right)\ \left(a+d\right)\ \left(a+2d\right)+d^4$$

$$=\left(a^2+3ad\right)\ \left(a^2+2ad+ad+2d^2\right)+d^4$$

$$=\left(a^2+3ad\right)\ \left(a^2+3ad+2d^2\right)+d^4$$

Let  $$t=a^2+3ad$$

$$=\left(t\right)\ \left(t+2d^2\right)+d^4$$

$$=t^2+2td^2+d^4$$

$$=\left(t+d^2\right)^2$$

Substituting  $$t=a^2+3ad$$

$$y=\left(a^2+3ad+d^2\right)^2$$.........(2)

From Eq (1) & (2)

We get, $$x+y=\left(a^2-d^2+ad\right)^2+\left(a^2+d^2+3ad\right)^2$$

Option (D) is correct.