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

Given,

$\sf{a=2+\sqrt{3} .}$

To find :-

$\sf{The\ value\ of:a^3+\dfrac{1}{a^3} .}$

Solution :-

First, let us calculate a³.

$\sf{a^3=(2+\sqrt{3} )^3}\\\\\sf{\dashrightarrow a^3=(2)^3 + (\sqrt{3} )^3 + 3 \times 2 \times \sqrt{3} (2+\sqrt{3} )}\\\\\sf{\dashrightarrow a^3=8+3\sqrt{3}+6\sqrt{3}(2+\sqrt{3} ) }\\\\\sf{\dashrightarrow a^3=8+3\sqrt{3}+(6\sqrt{3}\times 2)+(6\sqrt{3}\times\sqrt{3} ) }\\\\\sf{\dashrightarrow a^3=8+3\sqrt{3} +12\sqrt{3} +18}\\\\\sf{\dashrightarrow a^3=26+15\sqrt{3} }$

$\rule{130}2$

Now calculating 1/a³.

$\displaystyle{\sf{\dashrightarrow \frac{1}{a^3}=\frac{1}{26+15\sqrt{3}} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{a^3}=\frac{26-15\sqrt{3}}{(26+15\sqrt{3})(26-15\sqrt{3})} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{a^3}= \frac{26-15\sqrt{3}}{(26)^2-(15\sqrt{3})^2} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{a^3}= \frac{26-15\sqrt{3}}{676-675} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{a^3}= \frac{26-15\sqrt{3}}{1} }}$

$\displaystyle{\sf{\dashrightarrow \frac{1}{a^3}= 26-15\sqrt{3} }}$

$\rule{130}2$

Now,

$\displaystyle{\sf{a+ \frac{1}{a^3}=26+15\sqrt{3}+(26-15\sqrt{3}) }}\\\\\displaystyle{\sf{\dashrightarrow a+ \frac{1}{a^3}=26+15\sqrt{3}+26-15\sqrt{3}}}\\\\\boxed{\boxed{\displaystyle{\sf{\dashrightarrow a+ \frac{1}{a^3}=52}}}}$