Question

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If x=3+2√2 find the value of √x-1/√x)^3

Answer 1

Answer:

$:\implies\tt x = 3+2\sqrt{2}\\\\\\:\implies\tt x=1+2+2\sqrt{2}\\\\\\:\implies\tt x=(1)^2+(\sqrt{2})^2+(1 2 \sqrt{2})\\\\{\scriptsize\qquad\bf{\dag}\:\:\sf{(a^2+b^2+2ab)=(a+b)^2}}\\\\:\implies\tt x=(1+\sqrt{2})^2\\\\\\:\implies\boxed{\tt\sqrt{x}=(1+\sqrt{2})}\\\\\\\\:\implies\tt \dfrac{1}{\sqrt{x}}=\dfrac{1}{(1+\sqrt{2})}\\\\{\scriptsize\qquad\bf{\dag}\:\:\textsf{Rationalization of x.}}\\\\:\implies\tt \dfrac{1}{\sqrt{x}}=\dfrac{1}{(1+\sqrt{2})}\times\dfrac{(1-\sqrt{2})}{(1-\sqrt{2})}\\\\\\:\implies\tt \dfrac{1}{\sqrt{x}}= \dfrac{(1 - \sqrt{2})}{1 - 2}\\\\\\:\implies\tt \dfrac{1}{\sqrt{x}}= \frac{(1 - \sqrt{2})}{-\:1}\\\\\\:\implies\boxed{\tt\dfrac{1}{\sqrt{x}}=(\sqrt{2} - 1)}$

$\rule{160}{1}$

$\underline{\bigstar\:\:\textsf{According to the Question :}}$

$\dashrightarrow\sf\:\:\sqrt{x} - \dfrac{1}{\sqrt{x}}=(1+\sqrt{2}) - (\sqrt{2}-1)\\\\\\\dashrightarrow\sf\:\:\sqrt{x} - \dfrac{1}{\sqrt{x}}=1 + \sqrt{2} - \sqrt{2} + 1\\\\\\\dashrightarrow\tt\:\:\sqrt{x} -\dfrac{1}{\sqrt{x}} = 2\\\\\\\dashrightarrow\sf\:\:\bigg(\sqrt{x}-\dfrac{1}{ \sqrt{x}}\bigg)^3 = (2)^3\\\\\\\dashrightarrow \boxed{\red{\sf\:\: \large\bigg(\sqrt{x}-\dfrac{1}{\sqrt{x}}\bigg)^3 = 8}}$

Answer 2

Given:-

☛ x = 3 + 2√2

To find:-

☛ (√x - 1/√x)³

Solution:-

Here, x = 3 + 2√2

⇒ x = 1 + 2 + 2 × √2

⇒ x = (1)² + (√2)² + 2 × 1 × √2

Comparing with (a + b)² = a² + 2ab + b² we get:-

⇒ x = (1 + √2)²

⇒ √x = 1 + √2

Now let's find the value of 1/√x

⇒ 1/√x = 1/ (1 + √2)

⇒ 1/√x = (1 × √2 - 1)/(√2 + 1)(√2 - 1)

[∵ Multiplying both numerator & denominator by (√2 - 1) ]

⇒ 1/√x = (√2 - 1)/(√2)² - 1

[ ∵ (a² - b²) = (a + b)(a - b) ]

⇒ 1/√x = √2 - 1/1

⇒ 1/√x = √2 - 1

Now let's find value of (√x - 1/√x)³

⇒ (√x - 1/√x)³

⇒ [(√2 + 1) - (√2 - 1)]³

⇒ [√2 + 1 - √2 + 1]³

⇒ [2]³

⇒ 8

So,Your required answer = 8

Therefore,

Required value of (√x - 1/√x)³ = 8