Question

if a =2+√3 find the value of a³+1\a³​

Answer 1

Question -

In the above question ,

We have the value of -

a = 2 + √3

To Find -

We have to find the value of a³+ 1 / a³

Solution -

$\sf{a = 2 + \sqrt{3}} \\ \\ \sf{ \implies { \dfrac{1}{a} = \dfrac{ 1 }{ 2 + \sqrt{3}} }} \\ \\ \sf{ \implies { \dfrac{1}{a} = \dfrac{ 1 }{ 2 + \sqrt{3}} \times \dfrac{ 2 - \sqrt{3} }{ 2 - \sqrt{3} } }} \\ \\ \sf{ \implies { \dfrac{1}{a} = \dfrac{ 2 - \sqrt{3}} { 1 } = 2 - \sqrt{3} }} \\ \\ \sf{ \therefore { a + \dfrac{1}{a} = 4 }} \\ \\ \sf{ \bold { Cubing \: both \: sides \: - }} \\ \\ \sf{ \implies { a^3 + \dfrac{1}{ a^3 } + 3 a \times \dfrac{1}{a} ( a + \dfrac{1}{a} ) = 64 }} \\ \\ \sf{ \bold { But \: a + \dfrac{1}{a} = 4 }} \\ \\ \sf{ \implies { a^3 + \dfrac{1}{ a^3 } + 12 = 64 }} \\ \\ \sf{ \implies { a^3 + \dfrac{1}{a^3 } = 52 }}$

Hence , the required value is 52.

________________

Answer 2

Answer:

Given:

• a = 2 + $\sf \sqrt{3}$.

Find:

• Find the value of $\sf \dfrac{a^3 + 1}{a^3}$.

Step-by-Step-Calculation:

⇒ $\bold{\dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3}}}$

⇒ $\bold{\dfrac{1}{a} = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}}$

⇒ $\bold{\dfrac{1}{a} = \dfrac{2 - \sqrt{3}}{4 - 3}}$

⇒ $\bold{\dfrac{1}{a} = 2 - \sqrt{3}}$

⇒ $\bold{a + \dfrac{1}{a} = 2 + \sqrt{3} + 2 - \sqrt{3}}$

⇒ $\bold{a + \dfrac{1}{a} = 4}$

Cubing both the sides:

We know that:

⇒ $\bold{ (a + \dfrac{1}{a})^3 = (4^3)}$

⇒ $\bold{a^3 + \dfrac{1}{a^3} + 30 \times \dfrac{1}{a} (a + \dfrac{1}{a}) = \boxed{64}}$

⇒ $\bold{a^3 + \dfrac{1}{a^3} + 3 ((4)) = \boxed{64}}$

⇒ $\bold{ a^3 + \dfrac{1}{a^3} = \boxed{64 - 12}}$

• Note, in the first equation here (4^3) gives us the value 12 and the other side subtracting (64 with 12) we get 52.

.•. ${\sf{\underline{\boxed{\green{\sf{ a^3 + \dfrac{1}{a^3} = 52 }}}}}}$