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Algebra Level 3

For a real number x x , if x 4 + 1 x 4 = 47 x^4 + \dfrac1{x^4 }= 47 , find the value of x 3 + 1 x 3 x^3 + \dfrac1{x^3} .


The answer is 18.

Sharan Vishnu by sharan vishnu , 21, India

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1 solution

x 4 + 1 x 4 = 47 x^4 + \dfrac1{x^4 }= 47
( x 2 + 1 x 2 ) 2 − 2 = 47 (x^2+\dfrac{1}{x^2})^2-2=47
x 2 + 1 x 2 = 7 x^2+\dfrac{1}{x^2}=7
( x + 1 x ) 2 − 2 = 7 (x+\dfrac{1}{x})^2-2=7
x + 1 x = 3 x+\dfrac{1}{x}=3 ....... ( 1 ) (1)


→ x 3 + 1 x 3 x^3 + \dfrac1{x^3}
( x + 1 x ) × ( x 2 + 1 x 2 − 1 ) (x+\dfrac{1}{x})×(x^2+\dfrac{1}{x^2}-1)
( x + 1 x ) × [ ( x + 1 x ) 2 − 3 ] (x+\dfrac{1}{x})×[(x+\dfrac{1}{x})^2-3] [Now, substituting values From (1)]
3 × 6 = 18 3×6=\boxed{18}

2 Helpful 0 Interesting 2 Brilliant 0 Confused

Nice solution, Abhay :)

Swapnil Das - 5 years, 6 months ago

Note that you needed the assumption that x x is a real number, in order to select the positive square root. Otherwise, there are 4 possible solutions.

Calvin Lin Staff - 5 years, 6 months ago

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