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Algebra Level 2

If x x is a positive real number satisfying x 2 + 1 x 2 = 14 x^2 + \dfrac1{x^2} = 14 , find x 3 + 1 x 3 x^3 + \dfrac1{x^3} .


The answer is 52.

Yash Mehan by Yash Mehan , 19, India

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4 solutions

Atul Shivam
Jan 13, 2016

[ x 3 + ( 1 x ) 3 ] = ( x + 1 x ) ( x 2 − 1 + ( 1 x ) 2 ) . . . . . . . . . ( 1 ) [x^3+(\frac{1}{x})^3]= (x+\frac{1}{x})( x^2-1+(\frac{1}{x})^2).........(1)

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

( x + 1 x ) 2 = 16 (x+\frac{1}{x})^2=16 , Now ( x + 1 x ) = 4 (x+\frac{1}{x})=4

Hence put it in equation to get [ x 3 + ( 1 x ) 3 ] = 52 [x^3+(\frac{1}{x})^3]=\boxed{52}

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Why did you use ( x + 1 x ) 2 (x+\frac{1}{x})^2 ?

A Former Brilliant Member - 5 years, 4 months ago

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He's using the fact that the given expression x 2 + 1 x 2 x^2 + \frac {1}{x^2} is equivalent to ( x + 1 x ) 2 − 2 (x + \frac {1}{x})^2 - 2 . This allows him to calculate that x + 1 x = 14 + 2 = 4 x + \frac {1}{x} = \sqrt{14 + 2} = 4 from the given information, and then substitute this into his first equation (which consists of multiplying x + 1 x x + \frac {1}{x} by x 2 + 1 x 2 − 1 x^2 + \frac {1}{x^2} - 1 , giving the answer as 4 × ( 14 − 1 ) = 52 4 \times (14-1) = 52 and hence solving the question).

Luke Johnson-Davies - 5 years, 4 months ago
Daniel Ferreira
Jan 17, 2016

Da condição ( x 2 + 1 x 2 ) = 14 \left ( x^2 + \frac{1}{x^2} \right ) = 14 , tiramos que:

( x 2 + 1 x 2 ) = 14 ( x + 1 x ) 2 − 2 = 14 ( x + 1 x ) 2 = 16 ( x + 1 x ) = 4 \\ \left ( x^2 + \frac{1}{x^2} \right ) = 14 \\\\ \left ( x + \frac{1}{x} \right )^2 - 2 = 14 \\\\ \left ( x + \frac{1}{x} \right )^2 = 16 \\\\ \boxed{\left ( x + \frac{1}{x} \right ) = 4}

Com efeito,

( x 2 + 1 x 2 ) ⋅ ( x + 1 x ) = x 3 + x 2 x + x x 2 + 1 x 3 14 ⋅ 4 = x 3 + x + 1 x + 1 x 3 56 = ( x 3 + 1 x 3 ) + ( x + 1 x ) 56 = ( x 3 + 1 x 3 ) + 4 ( x 3 + 1 x 3 ) = 52 \\ \left ( x^2 + \frac{1}{x^2} \right ) \cdot \left ( x + \frac{1}{x} \right ) = x^3 + \frac{x^2}{x} + \frac{x}{x^2} + \frac{1}{x^3} \\\\ 14 \cdot 4 = x^3 + x + \frac{1}{x} + \frac{1}{x^3} \\\\ 56 = \left ( x^3 + \frac{1}{x^3} \right ) + \left ( x + \frac{1}{x} \right ) \\\\ 56 = \left ( x^3 + \frac{1}{x^3} \right ) + 4 \\\\ \boxed{\boxed{\left ( x^3 + \frac{1}{x^3} \right ) = 52}}

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Ahmed R. Maaty
Jan 25, 2016

( x 3 + 1 x 3 ) = x 2 + 1 x 2 + 2 ⋅ ( x 2 + 1 x 2 − 1 ) = 14 + 2 ⋅ ( 14 − 1 ) = 16 ⋅ 13 = 4 ⋅ 13 = 52 (x^3+\frac{1}{x^3})=\sqrt{x^2+\frac{1}{x^2}+2} \cdot (x^2+\frac{1}{x^2}-1)=\sqrt{14+2} \cdot (14-1)=\sqrt{16} \cdot 13=4 \cdot 13=52

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Tamás Lengyel
Jan 25, 2016

Adding the equations together we get:

Notice that:

From here we get:

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