Can the No $4\sqrt{4-2\sqrt {3}}+\sqrt{97-56\sqrt 3}$ be an Integer?

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SUMMARY

The expression $4\sqrt{4-2\sqrt{3}}+\sqrt{97-56\sqrt{3}}$ cannot be an integer. The first term simplifies to $4\sqrt{2-\sqrt{3}}$, which is not an integer, and the second term simplifies to $\sqrt{(8-4\sqrt{3})^2}$, also not yielding an integer. Therefore, the entire expression does not result in an integer value.

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Can the No :$4\sqrt{4-2\sqrt {3}}+\sqrt{97-56\sqrt 3}$ be an iteger ,if yes prove it if no then prove it again
 
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$4-2\sqrt{3} = 3 - 2\sqrt{3} +1 = (\sqrt{3} -1)^2$

$97-56\sqrt{3} = 49 - 2(28\sqrt{3}) + 48 = (7 - 4\sqrt{3})^2$

$4\sqrt{(\sqrt{3}-1)^2} + \sqrt{(7-4\sqrt{3})^2} = 4\sqrt{3}-4 + 7 -4\sqrt{3} = 3$
 
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very good ,excellent
 

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