Consider the following statements
I. Let ABCD be a parallelogram which is not a rectangle. Then, \( \Large 2 \left(AB^{2}+ BC^{2}\right)  \Large \ne  AC^{2} + BD^{2} \)
II. If ABCD is a rhombus with AB =4 cm, then \( \Large AC^{2} + BD^{2} = n^{3} \) for some positive integer n. Which of the above statements is/are correct?

Statement I ABCD is a parallelogram,

then

In \( \Large \triangle ABC, \)

\( \Large AC^{2} = AB^{2} + BC^{2} \) ...(i)

and in \( \Large \triangle CBD, \)

\( \Large BD^{2} = BC^{2} + CD^{2} \) ...(ii)

From Eqs. (i) and (ii), we get

\( \Large AC^{2} + BD^{2} = AB^{2}+BC^{2}+BC^{2}+CD^{2} \)

= \( \Large 2AB^{2} + 2BC^{2} \)

\( \Large AC^{2} + BD^{2} = 2 \left(AB^{2}+BC^{2}\right) \)

Statement II ABCD is a rhombus and diagonals AC and BD bisect each other.

Therefore, AO = OC and OB = OD

\( \Large AB^{2} = AO^{2} + OB^{2} \)

=> \( \Large \left(4\right)^{2} = \left(\frac{AC}{2}\right)^{2} + \left(\frac{BD}{2}\right)^{2} \)

\( \Large \therefore AC^{2} + BD^{2} = 64 = \left(4\right)^{3} \)

i.e., \( \Large n^{3} \)