root 8 + root 7 upon root 8 minus root 7 Y is equal to root 8 minus root 7 upon root 8 + root 7 find the value of x square + y square minus 3xy
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Hey friend,
Here is the answer you were looking for:

On rationalizing the denominator we get:

Using the identities:



On rationalizing the denominator we get:

Using the identities:



Putting the values:

Using same identities:

Hope this helps!!
If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.
@Mahak24
Thanks...
☺☺
Here is the answer you were looking for:
On rationalizing the denominator we get:
Using the identities:
On rationalizing the denominator we get:
Using the identities:
Putting the values:
Using same identities:
Hope this helps!!
If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.
@Mahak24
Thanks...
☺☺
DaIncredible:
thanx sir
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