Question

if x=3-2root2,find the value of x^3+1/x^3

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Answer 1

Given :-

$\sf{{x = 3 - 2\sqrt{2}}}$

To find :-

• $\sf{{{x}^{3}}}+$$\sf\dfrac{1}{{x}^{3}}$

Formula :-

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

$\implies\sf{{{x}^{3}}}+$$\sf\dfrac{1}{{x}^{3}}$=($\sf{{x}}+$$\sf\dfrac{1}{x}$)³$\sf{{-3}}$($\sf{{x+}}$$\sf\dfrac{1}{x}$)

Solution :-

As we know,

$\sf{{x = 3 - 2\sqrt{2}}}$

So,

$\sf\dfrac{1}{x}$=$\sf\dfrac{1}{3-2√2}$

Rationalizing Factor = 3 + 2√2

$=\sf\dfrac{1}{(3-2\sqrt{2})}$×$\sf\dfrac{(3+2\sqrt{2})}{(3+2\sqrt{2})}$

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

$=\sf\dfrac{(3+2\sqrt{2})}{9-8}$

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

Now,

$\sf{{x+}}$$\sf\dfrac{1}{x}$$\sf{{=(3-2\sqrt{2})+(3+2\sqrt{2})}}$

$\sf{{=3-2\sqrt{2}+3+2\sqrt{2}}}$

$\sf{{=3+3-2\sqrt{2}+2\sqrt{2}}}$

$\sf{{=6}}$

Now let us apply the Formula :

$\sf{{{x}^{3}}}+$$\sf\dfrac{1}{{x}^{3}}$=($\sf{{x}}+$$\sf\dfrac{1}{x}$)³$\sf{{-3}}$($\sf{{x+}}$$\sf\dfrac{1}{x}$)

$=\sf{{{(6)}^{3}}}$$\sf{{-3(6)}}$

$\sf{{=216-18}}$

$\sf{{=198}}$

★ Required Answer:

$\sf{{Value}}$$\sf{{of}}$$\sf{{{x}^{3}}}+$$\sf\dfrac{1}{{x}^{3}}$$\sf{{=198}}$