The Hard Equality

Algebra Level 3

If x 4 + 1 x 4 = 727 x^4 + \frac {1}{x^4} = 727 , find ∣ x 3 − 1 x 3 ∣ \left | x^3 - \frac{1}{x^3} \right |

( x x is a real number)


The answer is 140.

Jason Chrysoprase by Jason Chrysoprase , 18, Indonesia

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1 solution

Adding 2 to both sides of the equation, x 4 + 1 x 4 + 2 = 729 x^4+\frac{1}{x^4}+2=729 or ( x 2 + 1 x 2 ) 2 = 2 7 2 (x^2+\frac{1}{x^2})^2=27^2 or x 2 + 1 x 2 = 27 x^2+\frac{1}{x^2}=27 .

Now, subtracting 2 from both sides, x 2 + 1 x 2 − 2 = 25 x^2+\frac{1}{x^2}-2=25 or ( x − 1 x ) 2 = 5 2 (x-\frac{1}{x})^2=5^2 or x − 1 x = 5 x-\frac{1}{x}=5 .

x 3 − 1 x 3 = ( x − 1 x ) ( x 2 + 1 x 2 + x 3 × 1 x 3 ) = 5 × ( 27 + 1 ) = 5 × 28 = 140 x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+x^3 \times \frac{1}{x^3})=5 \times (27+1)=5 \times 28 =\boxed{140} .

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Nice solution

Keep up on the good work

Jason Chrysoprase - 5 years, 4 months ago

Actually, x 3 − 1 x 3 = ( x − 1 x ) 3 + 3 ( x − 1 x ) x^3 - \frac{1}{x^3} = (x -\frac{1}{x})^3 + 3(x -\frac{1}{x})

x 3 − 1 x 3 = 5 3 + 3 ( 5 ) x^3 - \frac{1}{x^3}=5^3 +3(5)

x 3 − 1 x 3 = 125 + 15 x^3 - \frac{1}{x^3}= 125 + 15

x 3 − 1 x 3 = 140 x^3 - \frac{1}{x^3} = 140

Jason Chrysoprase - 5 years, 4 months ago

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