Question

if x is equal to 9minus4root5 then find x cube plus 1 /x cube​

Answer 1

Answer:

x = 9 - 4 \sqrt{5}x=9−45

To find :

x {}^{3} + \frac{1}{x {}^{3} }x3+x31

Solution :

\begin{lgathered}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}\end{lgathered}x=9−45x1=9−451×9+459+45x1=(9)2−(45)29+45x1=81−809+45x1=9+45

Now,

\begin{lgathered}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\end{lgathered}x+x1=9−45+9+45x+x1=9+9x+x1=18

And, on cubing both sides.

\begin{lgathered}(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\end{lgathered}(x+x1)3=(18)3x3+x31+3(x+x1)=5826x3+x31+3×18=5826x3+x31+54=5826x3+x31=5826−54x3+x31=5778