2. If a = 3+2, 2 then find the value of a3+1/a3
Answers
Step-by-step explanation:
Given:
a = 3 + 2\sqrt{2}
2
To find, the value of a^{3} +\dfrac{1}{a^{3}}a
3
+
a
3
1
= ?
Solution:
∴ \dfrac{1}{a}
a
1
= \dfrac{1}{3 + 2\sqrt{2}}
3+2
2
1
Rationalising by denominator, we get
\dfrac{1}{a}
a
1
= \dfrac{1}{3 + 2\sqrt{2}}\times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}
3+2
2
1
×
3−2
2
3−2
2
⇒ \dfrac{1}{a}
a
1
= \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2}
3
2
−(2
2
)
2
3−2
2
⇒ \dfrac{1}{a}
a
1
= \dfrac{3 - 2\sqrt{2}}{9-8}
9−8
3−2
2
⇒ \dfrac{1}{a}
a
1
= 3 - 2\sqrt{2}
2
Using the algebraic identity:
(a+\dfrac{1}{a})^3=a^{3} +\dfrac{1}{a^{3}}+3(a) (\dfrac{1}{a})(a+\dfrac{1}{a})(a+
a
1
)
3
=a
3
+
a
3
1
+3(a)(
a
1
)(a+
a
1
)
∴ a^{3} +\dfrac{1}{a^{3}}=(a+\dfrac{1}{a})^3-3(a+\dfrac{1}{a})a
3
+
a
3
1
=(a+
a
1
)
3
−3(a+
a
1
)
Put a = 3 + 2\sqrt{2}
2
and \dfrac{1}{a}
a
1
= 3 - 2\sqrt{2}
2
, we get
a^{3} +\dfrac{1}{a^{3}}=(3 + 2\sqrt{2} +3 - 2\sqrt{2})^3-3(3 + 2\sqrt{2} +3 - 2\sqrt{2})a
3
+
a
3
1
=(3+2
2
+3−2
2
)
3
−3(3+2
2
+3−2
2
)
⇒ a^{3} +\dfrac{1}{a^{3}}=(6)^3-3(6)a
3
+
a
3
1
=(6)
3
−3(6)
⇒ a^{3} +\dfrac{1}{a^{3}}a
3
+
a
3
1
= 216 - 18
⇒ a^{3} +\dfrac{1}{a^{3}}a
3
+
a
3
1
= 198
∴ a^{3} +\dfrac{1}{a^{3}}a
3
+
a
3
1
= 198
Thus, if a = 3 + 2\sqrt{2}
2
, then a^{3} +\dfrac{1}{a^{3}}a
3
+
a
3
1
= 198.
Answer:
226/15
Step-by-step explanation:
a =3+2
=5
Now,
a3+1/a3
= 5x3+1/5x3
= 15+1/15
= 225/15+1/15
= 226/15