If $$a^4+b^4=17$$​ and $$a+b=1$$​, then the value of $$a^2b^2-2ab$$​ is:

Correct option is C

​Given:​​

We have two equations:
​$$a^4 + b^4 = 17$$​​
a + b = 1 

Formula Used:

​$$a^2 + b^2 = (a + b)^2 - 2ab$$​​

Solution:
​$$a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \\ \ \\a^2 + b^2 = (a + b)^2 - 2ab = 1^2 - 2ab = 1 - 2ab \\ \ \\a^4 + b^4 = (1 - 2ab)^2 - 2a^2b^2 \\ \ \\(1 - 2ab)^2 - 2a^2b^2 = 17 \\ \ \\1 - 4ab + 4a^2b^2 - 2a^2b^2 = 17 \\ \ \\1 - 4ab + 2a^2b^2 = 17 \\ \ \\2a^2b^2 - 4ab + 1 - 17 = 0 \\ \ \\2a^2b^2 - 4ab - 16 = 0 \\ \ \\a^2b^2 - 2ab - 8 = 0\\\ \\a^2b^2 - 2ab = 8$$

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