Answers

2014-05-18T16:59:36+05:30
Okay, this is a good question :-
It is given that  x^{2} +  \frac{1}{ x^{2} } = 18 .....................(1)
We also know that  (x- \frac{1}{x} )^{2} =  x^{2}+ \frac{1}{ x^{2} } + 2  [using identities]
So, we can rewrite (1) as (x- \frac{1}{x} )^{2} +2 = 16 +2 
=> (x- \frac{1}{x} )^{2} = 16 
now, x- \frac{1}{x}  \sqrt{16} = 4  
we know that  a^{3} -  b^{3} = (a-b)( a^{2} + ab +  b^{2}    )
So,x^{3} - \frac{1}{ x^{3} } = (x- \frac{1}{x})( x^{2} + \frac{1}{ x^{2} } + 1) [1 because ab = x* \frac{1}{x} = 1 ]
Now, substituting values of ( x^{2} + \frac{1}{ x^{2} } + 1) and (x- \frac{1}{x}) 
We Get :-
x^{3} - \frac{1}{ x^{3} } = 4* (18 + 1) = 4*19 = 76

Please recheck if there is any mistake in working :-)


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The Brainliest Answer!
2014-05-18T19:41:57+05:30
If x² + 1/x² = 18
 using identity (a+b)² = a² +2ab +b²
 (x)² + 2×x×1/x =18
 x² + 2 + 1/x² = 18
x² + 1/x² = 18 - 2
x² + 1/x² = 16
{square rooting both sides}
\sqrt {x^{2} + \frac{1}{x^{2}}=  \sqrt{16}  
x +   \frac{1}{x} = 4
{cubing both the sides}
{ x -  \frac{1}{x} }³ = 4³
 x³ -  \frac{1}{x^{3}} - 3×x×  \frac{1}{x} { x +  \frac{1}{x} } =64
x³ -  \frac{1}{x^{3}} - 3{4} = 64
x³ -  \frac{1}{x^{3}} - 12 = 64
x³ -  \frac{1}{x^{3}} = 64 + 12
therefore x³ +  \frac{1}{x^{3}} = 76
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