Question

if a=3+2√2 find the value of a3+1/a3​

Answer 1

Answer:

$(12 \sqrt{2 } + 18) \div 3 + 2 \sqrt{2}$

Step-by-step explanation:

${a}^{3} + 1 \div {a}^{3} =$

$3 + 2 \sqrt{2} + 1 \div 3 + 2 \sqrt{2}$

$(12 \sqrt{2 } + 18) \div 3 + 2 \sqrt{2}$

Answer 2

Given:

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

To find, the value of $a^{3} +\dfrac{1}{a^{3}}$ = ?

Solution:

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

Rationalising by denominator, we get

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

⇒ $\dfrac{1}{a}$ = $\dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2}$

⇒ $\dfrac{1}{a}$ = $\dfrac{3 - 2\sqrt{2}}{9-8}$

⇒ $\dfrac{1}{a}$ =  3 - 2$\sqrt{2}$

Using the algebraic identity:

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

∴ $a^{3} +\dfrac{1}{a^{3}}=(a+\dfrac{1}{a})^3-3(a+\dfrac{1}{a})$

Put a = 3 + 2$\sqrt{2}$ and $\dfrac{1}{a}$ =  3 - 2$\sqrt{2}$, we get

$a^{3} +\dfrac{1}{a^{3}}=(3 + 2\sqrt{2} +3 - 2\sqrt{2})^3-3(3 + 2\sqrt{2} +3 - 2\sqrt{2})$

⇒ $a^{3} +\dfrac{1}{a^{3}}=(6)^3-3(6)$

⇒ $a^{3} +\dfrac{1}{a^{3}}$ = 216 - 18

⇒ $a^{3} +\dfrac{1}{a^{3}}$ = 198

∴ $a^{3} +\dfrac{1}{a^{3}}$ = 198

Thus, if a = 3 + 2$\sqrt{2}$, then $a^{3} +\dfrac{1}{a^{3}}$ = 198.