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Algebra and Applications Test 2

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Question : 10 of 40

Marks: +1, -0

If

\left(x+\frac{1}{x}\right)^2=3

then the value of

(x^{72}+x^{66}+x^{54}+x^{36}+x^{24}+x^{6}+1)

is:

1
2
3
4

Solution: 👈: Video Solution

Concept:

The given condition reduces to

x^6=-1

, which lets us simplify all high powers of

x

.

Explanation:

Given

\left(x+\frac{1}{x}\right)^2=3

, we get

x+\frac{1}{x}=\pm\sqrt{3}

.

Cube both sides:

\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)

Since

\left(x+\frac{1}{x}\right)^3=3\left(x+\frac{1}{x}\right)

, we have:

x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=3\left(x+\frac{1}{x}\right)

So,

x^3+\frac{1}{x^3}=0

.

Multiply by

x^3

:

x^6+1=0

, hence

x^6=-1

.

Let

y=x^6=-1

.

The expression becomes:

y^{12}+y^{11}+y^9+y^6+y^4+y+1

Substitute

y=-1

:

(-1)^{12}+(-1)^{11}+(-1)^9+(-1)^6+(-1)^4+(-1)+1

=1-1-1+1+1-1+1=1

Answer:

1

Correct option: A.

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