Question

if x is equal to 9 minus 4 root 5 then find x cube + 1 upon x cube

Answer 1

When x=9-4√5
Then we have to find 1/x which should be equals to 9+4√5

1/x = 9+4√5

Then (x + 1/x )whole cube = x cube + 1 / x cube + 3 ( x + 1/x)

(9-4√5+9+4√5) cube = x cube + 1/x cube + 3(9-4√5+9+4√5)

(18) cube = x cube + 1/ x cube + 3(18)

5832= x cube + 1/x cube+54

x cube + 1/x cube = 5832 - 54

So, x cube +1/x cube = 5778

Answer 2

Hey mate!

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Given :

$x = 9 - 4 \sqrt{5}$

To find :

$x {}^{3} + \frac{1}{x {}^{3} }$

Solution :

$x = 9 - 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5}} \times \frac{9 + 4 \sqrt{5}}{9 + 4 \sqrt{5}} \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{(9) {}^{2} - (4 \sqrt{5} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 + 4 \sqrt{5}$

Now,

$x + \frac{1}{x} = 9 - \cancel{4 \sqrt{5} } + 9 + \cancel{4 \sqrt{5}} \\ \\ x + \frac{1}{x} = 9 + 9 \\ \\ x + \frac{1}{x} = 18$
And, on cubing both sides.

$(x + \frac{1}{x} ) {}^{3} = (18){}^{3} \\ \\ x{}^{3} + \frac{1}{x{}^{3} } + 3(x + \frac{1}{x} ) = 5826 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } + 3 \times 18 = 5826 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } + 54 = 5826 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } = 5826 - 54 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } = 5778$


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Thanks for the question !

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