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Answer 1

Question:

If x + 1/x = ³√(4 + 2√2) + ³√(4 - 2√2), then the value of [√{(x³ + 1/x³) - 3(x + 1/x) + 1}/3] is

A. 0

B. 1

C. 2

D. 3

Answer:

B. 1

Step-by-step explanation:

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{4 + 2 \sqrt{2}} \: + \: ^{3} \sqrt{4 - 2 \sqrt{2} }}}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{4} \: + \: ^{3} \sqrt{4 }}}$

$\sf{\rightarrow\:x+\frac{1}{x}=3.17}$

$\sf{\rightarrow{\purple{3}\:(approx.)}}$

$\rule{70mm}{1pt}$

OR

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{4 + 2 \sqrt{2}} \: + \: ^{3} \sqrt{4 - 2 \sqrt{2} }}}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{4} + \: ^{3}\sqrt{2 \sqrt{2}}} \: + \: ^{3} \sqrt{4} \: + \: ^{3}\sqrt{2 \sqrt{2}}}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \:1.58 + 1.78 + 1.58 - 1.78}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \:3.16}$

$\sf{\rightarrow{\purple{3}\:(approx.)}}$

$\rule{70mm}{1pt}$

OR

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{4 + 2 \sqrt{2}} \: + \: ^{3} \sqrt{4 - 2 \sqrt{2} }}}$

Let's say 4 = a and 2√2 = b.

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{a + b} \: + \: ^{3} \sqrt{a - b }}}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: ^{3}{\sqrt{a} + \: ^{3}{ \sqrt{b} \: + \: ^{3} \sqrt{a} - \: ^{3} \sqrt{b}}}}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: a^{ \frac{1}{3}} \: + \: a^{ \frac{1}{3}}}$

$\sf{ \rightarrow \: x+ \frac{1}{x} = \: 4^{ \frac{1}{3}} \: + \: 4^{ \frac{1}{3}} \: \: \: (As,\: a = 4)}$

$\sf{\rightarrow\:x+\frac{1}{x}=3.16}$

$\sf{\rightarrow{\purple{3}\:(approx.)}}$

Now,

$\implies\:\sf\purple{x+1/x=3}$

Do cube on both sides,

$\implies\:\sf{(x+1/x)^{3} = {3}^{3} }$

$\implies\:\sf{x^3+(1/x)^3+3(x)(1/x)=27}$

$\implies\:\sf{x^3+(1/x)^3=27-3}$

$\implies\:\sf\purple{x^3+(1/x)^3=24}$

From above we have value of x + 1/x = 3 and x³ + 1/x³ = 24. Substitute value of x + 1/x and x³ + 1/x³ in [√{(x³ + 1/x³) - 3(x + 1/x) + 1}/3].

$\sf\Rightarrow\:{ \dfrac{ \sqrt{( {x}^{3} + \frac{1}{ {x}^{3} }) \: -3(x + \frac{1}{x})+1}}{3} \:=\:\dfrac{ \sqrt{(24 - 3(3) + 1)} }{3}}$

$\sf\Rightarrow \: {\dfrac{ \sqrt{( {x}^{3} + \frac{1}{ {x}^{3} }) \: -3(x + \frac{1}{x})+1}}{3}= \dfrac{ \sqrt{(24 - 9+ 1)} }{3}}$

$\sf\Rightarrow \: {\dfrac{ \sqrt{( {x}^{3} + \frac{1}{ {x}^{3} }) \: -3(x + \frac{1}{x})+1}}{3}= \dfrac{ \sqrt{(24 - 8)} }{3}}$

$\sf\Rightarrow \: {\dfrac{ \sqrt{( {x}^{3} + \frac{1}{ {x}^{3} }) \: -3(x + \frac{1}{x})+1}}{3}=\dfrac{ \sqrt{16)} }{3}}$

$\sf\Rightarrow\:\dfrac{ \sqrt{( {x}^{3} + \frac{1}{ {x}^{3} }) \: -3(x + \frac{1}{x})+1}}{3}=\dfrac{4}{3}$

$\sf\Rightarrow\:{\dfrac{ \sqrt{( {x}^{3} + \frac{1}{ {x}^{3} }) \: -3(x + \frac{1}{x})+1}}{3}=1.33}$

$\sf\Rightarrow\:\purple{1} \:(approx.)$

Therefore, the value of [√{(x³ + 1/x³) - 3(x + 1/x) + 1}/3] is 1.