if x=2+³√, find the value of x³+1/x³
Answers
Step-by-step explanation:
Hi there !
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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} \\ \\ \implies2 + \cancel{\sqrt{3}}+ 2 - \cancel {\sqrt{3}} \\ \\ \implies2 + 2 \\ \\ \implies4\end{gathered}
x+
x
1
⟹2+
3
+2−
3
⟹2+2
⟹4
So, on cubing both sides, we have
\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 \\ \\ 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
=64−12
x
3
+
x
3
1
=52
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Thanks for the question !
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