I was helping my son with homework and was able to solve this problem (verified answer with textbook) by considering that there are total $8!$ ways of arranging the letters. Now considering that no vowel is next to the other, we have $5!$ ways of arranging 5 consonants and with 6 slots available for vowels we choose $6 \choose 3$ ways of selecting slots for 3 vowels. And again the 3 vowels can be arranged in $3!$ ways. So the answer is $8! - 5! $ $6 \choose 3$ $3! = 25920$. This matches text book answer.
Now follows my son's chain of thought and I am unable to spot the flaw in this. Please help.
We can choose 2 out of 3 vowels in $3 \choose 2$ ways and the can flipped around in 2 ways. The remaining letters are 6 (5 consonants, 1 vowel) and they can be arranged in $6!$ ways. There are seven slots among the six remaining letters and the pair of two vowels next to each other can be placed in any one of them. So in all we have $3 \choose 2$ $2$ $6!$ $7$ = 30240.