If x = 2-√3 Find the value of (x+1/x)3 ans is -24√3
Answers
Step-by-step explanation:
Given :
x = 2 - \sqrt{3}x=2−
3
To find :
x {}^{3} + \frac{1}{x{}^{3}}x
3
+
x
3
1
Solution :
\begin{gathered}x = 2 - \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 + \sqrt{3} \end{gathered}
x=2−
3
x
1
=
2−
3
1
×
2+
3
2+
3
x
1
=
(2)
2
−(
3
)
2
2+
3
x
1
=
4−3
2+
3
x
1
=2+
3
Now,
\begin{gathered}x + \frac{1}{x} = 2 - \cancel {\sqrt{3}} + 2 + \cancel {\sqrt{3}} \\ \\ x + \frac{1}{x} = 2 + 2 \\ \\ x + \frac{1}{x} = 4\end{gathered}
x+
x
1
=2−
3
+2+
3
x+
x
1
=2+2
x+
x
1
=4
And,
On cubing both sides.
\begin{gathered}(x + \frac{1}{x} ){}^{3} = (4) {}^{3} \\ \\ x {}^{3} + \frac{1}{x{}^{3} } + 3(x + \frac{1}{x} ) = 64 \\ \\ x {}^{3} + \frac{1}{x{}^{3} } + 3 \times 4 = 64 \\ \\ x {}^{3} + \frac{1}{x{}^{3} } + 12 = 64 \\ \\ x {}^{3} + \frac{1}{x{}^{3} } = 64 - 12 \\ \\ \boxed{ \bold{ x {}^{3} + \frac{1}{x{}^{3} } = 52}}\end{gathered}
(x+
x
1
)
3
=(4)
3
x
3
+
x
3
1
+3(x+
x
1
)=64
x
3
+
x
3
1
+3×4=64
x
3
+
x
3
1
+12=64
x
3
+
x
3
1
=64−12
x
3
+
x
3
1
=52
_______________________
Thanks for the question!