If a2+1/a=5 then find the value of a3+1/a3
Answers
⚫ Given :- a = 2 + √3
⚫ To Find :- a³ + 1/a³ = ?
⚫ Solution :- a = 2 + √3
\begin{lgathered}\frac{1}{a} = \frac{1}{2 + \sqrt{3} } \\ \\ \frac{1}{a} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{a} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{a} = 2 - \sqrt{3} \\ \\ a + \frac{1}{a} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ a + \frac{1}{a} = 4 \\ \\ take \: \: cube \: \: both \: \: side \: \: \\ \\ {(a + \frac{1}{a} )}^{3} = {4}^{3} \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3a \times \frac{1}{a} (a + \frac{1}{a} ) = 64 \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } + 3(4) = 64 \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } = 64 - 12 \\ \\ {a}^{3} + \frac{1}{ {a}^{3} } = 52\end{lgathered}
a
1
=
2+
3
1
a
1
=
2+
3
1
×
2−
3
2−
3
a
1
=
4−3
2−
3
a
1
=2−
3
a+
a
1
=2+
3
+2−
3
a+
a
1
=4
takecubebothside
(a+
a
1
)
3
=4
3
a
3
+
a
3
1
+3a×
a
1
(a+
a
1
)=64
a
3
+
a
3
1
+3(4)=64
a
3
+
a
3
1
=64−12
a
3
+
a
3
1
=52