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registerincog
The expression \((\sqrt{8 +\sqrt{63}} +\sqrt{8 -\sqrt{63}})^2\) =

A) 20
B) 19
C) 18
D) 17
E) 16

Since the given expression is in the form of (x + y)^2, we can use the fact that (x + y)^2 = x^2 + y^2 + 2xy. Thus:

x^2 = [√(8 + √63)]^2 = 8 + √63

y^2 = [√(8 - √63)]^2 = 8 - √63

2xy = 2[√(8 + √63)][√(8 - √63)] = 2√(8^2 - (√63)^2) = 2√(64 - 63) = 2

Thus, the final answer is 8 + √63 + 8 - √63 + 2 = 18.

Answer: C
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Deconstructing the Question

We are given

\(\left(\sqrt{8+\sqrt{63}}+\sqrt{8-\sqrt{63}}\right)^2\)

Use the identity

\((a+b)^2=a^2+b^2+2ab\)

Step-by-step

Let

\(a=\sqrt{8+\sqrt{63}}, \quad b=\sqrt{8-\sqrt{63}}\)

Then

\(a^2=8+\sqrt{63}\)

\(b^2=8-\sqrt{63}\)

So

\(a^2+b^2=16\)

Now compute

\(ab=\sqrt{(8+\sqrt{63})(8-\sqrt{63})}\)

\(=(8+\sqrt{63})(8-\sqrt{63})=64-63=1\)

Thus

\(ab=1\)

So

\((a+b)^2=16+2(1)=18\)

Answer: C
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I have one silly doubt and not able to wrap my mind on this..
When we are squaring like below, why we are not applying a^2+b^2+2ab rule again..?

x^2 = [√(8 + √63)]^2 = 8 + √63
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Brindac2
I have one silly doubt and not able to wrap my mind on this..
When we are squaring like below, why we are not applying a^2+b^2+2ab rule again..?

x^2 = [√(8 + √63)]^2 = 8 + √63

Because here you are not squaring 8 + √63. You are squaring √(8 + √63), and square and square root cancel each other, so [√(8 + √63)]^2 = 8 + √63.

The rule a^2 + b^2 + 2ab is used only when you square a sum like (a + b)^2. Here the entire sum 8 + √63 is inside one square root, so that expansion does not apply.
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