If two natural numbers a and b are such that $$a^2-b^2=17,a>b$$​ ，then a possible value of $$a^2+b^2-ab$$​ will be

Correct option is C

Given:
​$$a^2 - b^2 = 17$$​​
a and b are natural numbers, a > b
Formula Used:
​$$a^2 - b^2 = (a-b)(a+b)$$​​
Solution:
Express the given equation using the algebraic identity:
(a-b)(a+b) = 17
Since 17 is a prime number and a, b are natural numbers with a > b, the only factors are 1 and 17.
Thus, a - b = 1 and a + b = 17.
Adding the two equations:
2a = 18➔ a = 9
Subtracting the two equations:
2b = 16 ➔ b = 8
We need to find the value of $$a^2 + b^2 - ab.$$​​
Substitute the values of a and b:
​$$9^2 + 8^2 - (9 × 8) = 81 + 64 - 72$$​​
Result = 145 - 72 = 73
Final Answer
So the correct answer is (c)