Question

If a= 2+√3,find the value of a3+1/a3

Answer 1

Answer:

Given:

• a = 2+√3
• to find a³ + 1/a³

Solving Question:

    we are given the value of 'a' to find a³ + 1/a³  first, we have to find 1/a and then cube it.

Solution:

   $a = 2+\sqrt{3}\\\\to\:find\:\dfrac{1}{a}\\\\\implies \dfrac{1}{2+\sqrt{3}}\\Rationalise\:denominator\\\\\implies\dfrac{1}{2+\sqrt{3}}*\dfrac{2-\sqrt{3}}{2-\sqrt{3}}\implies \dfrac{2-\sqrt{3}}{4-3}\:( (a-b)(a+b)=a^2-b^2)\\\\or,=\dfrac{2-\sqrt{3}}{1} \:\implies 2-\sqrt{3}$

Hence, a = 2+√3 and 1/a = 2-√3

Then, cube it

⇒ a³ + 1/a³

⇒ (2+√3)³ + (2-√3)³

⇒ 2³ + (√3)³ +3*2*√3(2+√3) + 2³ - (√3)³ - 3*2*√3 (2-√3) [ ∵ (x+y)³ = x³ +y³+3xy(x+y) and (x-y)³ = x³ -y³ -3xy(x-y) ]

cancel +(√3)³ and -(√3)³

⇒ 8 +6√3(2+√3) +8- 6√3(2-√3)

⇒16 + 12√3 +18 -12√3+18

⇒ 16 + 18+18

⇒ 52

Therefore, the value of a³ + 1/a³ = 52

Answer 2

a=2+√3

$\frac{1}{a} = \frac{1}{2 + \sqrt{3} }$

Rationalise it with 2+√3

$\frac{1 \times 2 - \sqrt{3} }{(2 + \sqrt{3}) (2 - \sqrt{3} )} \\ \\ \frac{2 - \sqrt{3} }{ {2}^{2} - \sqrt{3} {}^{2} } \\ \\ \frac{2 - \sqrt{3} }{4 - 3} \\ \\ = 2 - \sqrt{3}$

$a = 2 + \sqrt{3} \\ \\ \frac{1}{a} = 2 - \sqrt{3}$

Now

.

${a}^{ 3 } + \frac{1}{ {a}^{3} }$

As we know

(a+b)³=a³+b³+3ab(a+b)

Here let a=2+√3

b=2-√3

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

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

${a}^{ 3} + \frac{1}{ {a}^{3} } = 26 + 15 \sqrt{3} + 26 - 15 \sqrt{3} \\ \\ = 26 + 26 \\ \\ = 52$

$\underline{\mathbf{\huge{{a}^{3}+\frac{1}{{a}^{3}}=52}}}$