0
votes

When $2x + \frac{2}{x} = 3,$ then the value of $x^{3} + \frac{1}{x^{3}} + 2$ is :

- $7/2$
- $7/8$
- $8/7$
- $2/7$

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0
votes

When $2x + \frac{2}{x} = 3,$ then the value of $x^{3} + \frac{1}{x^{3}} + 2$ is :

- $7/2$
- $7/8$
- $8/7$
- $2/7$

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0
votes

Given that: $2x+\frac{2}{x}=3$

$\implies 2(x+1/x)=3$

$\implies (x+\frac{1}{x})=\frac{3}{2}…….(i)$

cubing eq (i) we get:

$\implies x^3+\frac{1}{x^3}+3*x*\frac{1}{x}(x+\frac{1}{x})=\frac{27}{8}$

$\implies x^3+\frac{1}{x^3}+3*\frac{3}{2}=\frac{27}{8}$

$\implies x^3+\frac{1}{x^3}=\frac{27}{8}-\frac{9}{2}$

Adding $+2$ in both side we get:

$\implies x^3+\frac{1}{x^3}+2 =\frac{27}{8}-\frac{9}{2}+2$

$\implies x^3+\frac{1}{x^3}+2=\frac{27-36+16}{8}=\frac{7}{8}$

Option (B) is correct.

$\implies 2(x+1/x)=3$

$\implies (x+\frac{1}{x})=\frac{3}{2}…….(i)$

cubing eq (i) we get:

$\implies x^3+\frac{1}{x^3}+3*x*\frac{1}{x}(x+\frac{1}{x})=\frac{27}{8}$

$\implies x^3+\frac{1}{x^3}+3*\frac{3}{2}=\frac{27}{8}$

$\implies x^3+\frac{1}{x^3}=\frac{27}{8}-\frac{9}{2}$

Adding $+2$ in both side we get:

$\implies x^3+\frac{1}{x^3}+2 =\frac{27}{8}-\frac{9}{2}+2$

$\implies x^3+\frac{1}{x^3}+2=\frac{27-36+16}{8}=\frac{7}{8}$

Option (B) is correct.

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