Question

if x= 9+4√5 then find= x^3+1/x^3.

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = 9 + 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \\ \\ on \: rationalizing \: we \: get \\ \\ \frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a }^{2} - {b}^{2} \\ \\ \frac{1}{x} = \frac{9 - 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5} )}^{2} } \\ \\ \frac{1}{x} = \frac{9 - 4 \sqrt{5} }{81 - 16 \times 5} \\ \\ \frac{1}{x} = \frac{9 - 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 - 4 \sqrt{5} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } \\ \\ using \: the \: identities \\ {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) \\ {(a - b)}^{3} = {a}^{3} - {b}^{3} - 3ab(a - b) \\ \\ {(9 + 4 \sqrt{5} )}^{3} + (9 - 4 \sqrt{5} )^{3} \\ \\ = ( {(9)}^{3} + {(4 \sqrt{5}) }^{3} + 3 \times 9 \times 4 \sqrt{5} (9 + 4 \sqrt{5} ) \\ + ( {(9)}^{3} - {(4 \sqrt{5} )}^{3} - 3 \times 9 \times 4 \sqrt{5} (9 - 4 \sqrt{5} ) \\ \\ = (729 + 320 \sqrt{5} + 108 \sqrt{5} (9 + 4 \sqrt{5} ) \\ + (729 - 320 \sqrt{5} - 108 \sqrt{5} (9 - 4 \sqrt{5} ) \\ \\ = (729 + 320 \sqrt{5} + 108 \sqrt{5} \times 9 + 108 \sqrt{5} \times 4 \sqrt{5} ) \\ + (729 - 320 \sqrt{5} - 108 \sqrt{5} \times 9 + 108 \sqrt{5} \times 4 \sqrt{5} ) \\ \\ = (729 + 320 \sqrt{5} + 972 \sqrt{5} + 2160) \\ + (729 - 320 \sqrt{5} - 972 \sqrt{5} + 2160) \\ \\ = 729 + 320 \sqrt{5 } + 972 \sqrt{5} + 2160 \\ + 729 - 320 \sqrt{5} - 972 \sqrt{5} + 2160 \\ \\ = 729 + 729 + 2160 + 2160 \\ \\ = 5778$

Hope this helps!!!!

@Mahak24

Thanks...
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