If x is not equal to zero, and x+1/x = 3

Question

If x is not equal to zero, and x+1/x = 3, then what is the value of x^6 + (1/x)^6?

A. 320
B. 322
C. 324
D. 326
E. 328

pushpitkc · Answer

Given:   ( x + \frac{1}{x} ) = 3

Squaring on both sides, we get  (x + \frac{1}{x})^2 = 3^2  ->  x^2 +(\frac{1}{x^2}) + 2 = 9  ->  x^2 +(\frac{1}{x^2}) = 7

Applying cube on both sides, we get  (x^2 +(\frac{1}{x^2}))^3 = 7^3

Formula used:    (a + b)^3 = a^3 + b^3 + 3ab(a + b)

The equation simplifies to  (x^2)^3 + (\frac{1}{x^2})^3 + 3*x^2*\frac{1}{x^2}(x^2 +(\frac{1}{x^2})) = 343

Therefore, the value of  x^6 + (\frac{1}{x^6}) + 3*7 = 343  ->  x^6 + (\frac{1}{x^6}) = 343 - 21 = 322  (Option B)

Bunuel · Answer

chandra004 · Answer

My approach(I found it easy with squaring and proceeding):-
Given: x+1/x = 3
=> x^2 + (1/x)^2 + 2*x*(1/x) = 9
=> x^2 + (1/x)^2 = 7......(1)

Now, square eqn 1:-
x^4 + (1/x)^4 = 47.....(2)

Now multiply eqn 1 & 2:-
=> x^6 + (1/x)^2 + x^2 + (1/x)^6 = 329;
=> x^6 + 7 + (1/x)^6 = 329.....got "7" using eqn 1
=> x^6 + (1/x)^6 = 322

Hence, B is our answer.

Abhishek009 · Answer

Possible, but  pushpitkc  has shown the fastest possible approach for exam hall situation, it is best to adopt that method.

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