If x + ≤ft[ 1/x ] = p, then what is x^6 + ≤ft[ 1/X^6 ] equal to?

Correct Answer is : p6−6p4+9p2−2p^6 - 6p^4 + 9p² - 2p6−6p4+9p2−2

Correct Answer is : p6−6p4+9p2−2p^6 - 6p^4 + 9p² - 2p6−6p4+9p2−2

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RRB NTPC Mains or Stage 2 Model Paper 5 with solutions

Question : 16 of 120

Marks: +1, -0

x + \left[ \frac{1}{x} \right] = p

x^6 + \left[ \frac{1}{X^6} \right]

equal to?

Solution:

Concept:

Use the identities

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

and

x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2

Explanation:

Given

x+\frac{1}{x}=p

First compute

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

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

Now square this result:

x^6+\frac{1}{x^6}=\left(x^3+\frac{1}{x^3}\right)^2-2.

Substitute

x^3+\frac{1}{x^3}=p^3-3p

x^6+\frac{1}{x^6}=(p^3-3p)^2-2.

Simplify:

=p^6-6p^4+9p^2-2.

Answer:

x^6+\frac{1}{x^6}=p^6-6p^4+9p^2-2

Hence, option D is correct.

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