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Explanation:

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Given, BC = 5 and BE = 3

In △ABD ⇒ AB2 = AD2 + BD2    ...(1)
In △AED ⇒ AE2 = AD2 + DE2    ...(2)

(1) - (2) 
⇒ AB2 - AE2 = BD2 - DE2 
⇒ AB2 - AE2 = x2 - (3 - x)2 
⇒ AB2 - AE2 = x2 - 9 - x2 + 6x 
⇒ AB2 - AE2 = - 9 + 6x 

We need to find AB2 - AE2 + 6CD
= (-9 + 6x) + 6 × (5 - x)
= -9 + 6x + 30 - 6x
= 21

Hence, option (e).

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