Question

If $\frac{x^2+1}{x}=3\frac{1}{3}$ and x > 1 , then find $x^3-\frac{1}{x^3}$.

• A. $6\frac{26}{27}$
• B. $26\frac{26}{27}$
• C. $13\frac{26}{27}$
• D. $42\frac{26}{27}$

Answer 1

Answer: (B)

$26\frac{26}{27}$

$\frac{x^2+1}{x}=3\frac{1}{3}$

$=>x+\frac{1}{x}=\frac{10}{3}$

Squaring both sides,

$(x+\frac{1}{x}{)}^2=(\frac{10}{3}{)}^2$

$=>x^2+(\frac{1}{x}{)}^2+2(x\left)\right(\frac{1}{x})=\frac{100}{9}$

$=>x^2+(\frac{1}{x}{)}^2=\frac{100}{9}-2$

$=>x^2+(\frac{1}{x}{)}^2=\frac{82}{9}$

Now, $(x-\frac{1}{x}{)}^2=x^2+(\frac{1}{x}{)}^2-2\left(x\right)\left(\frac{1}{x}\right)$

$=(\frac{82}{9}-2)$

$=\left(\frac{64}{9}\right)$

$(x-\frac{1}{x})=\sqrt{(\frac{64}{9}})$

$=\left(\frac{8}{3}\right)$

$=2\frac{2}{3}$

Now, $(x^3-\frac{1}{x^3})=(x-\frac{1}{x})\left[\right(x^2+(\frac{1}{x}{)}^2+(x\left)\right(\frac{1}{x}\left)\right]$

$=\left(\frac{8}{3}\right)(\frac{82}{9}+1)$

$=\left(\frac{8}{3}\right)\left(\frac{91}{9}\right)$

$=\frac{728}{27}$

$=26\frac{26}{7}$