Question

If $x^4+\frac{1}{x^4}=\frac{257}{16}$ then find $\frac{8}{13}(x^3+\frac{1}{x^3})$, where x>0.

• A. 5
• B. 8
• C. 4
• D. 6

Answer 1

Answer:

A. 5

Given:

$x^4 + \frac{1}{x^4} = \frac{257}{16}$

Formula:

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

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

Calculation:

$(x^2+\frac{1}{x^2})^2=x^4+\frac{1}{x^4}+2$

⇒$(x^2+\frac{1}{x^2})^2=(\frac{257}{16})+2$

⇒ $(x^2+\frac{1}{x^2})^2=\frac{(257+32)}{16}$

⇒$(x^2+\frac{1}{x^2})=\sqrt(\frac{289}{16})$

⇒$(x^2+\frac{1}{x^2})=\frac{17}{4}$

Again,

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

⇒$(x+\frac{1}{x})^2=(\frac{17}{4})+2$

 

⇒$(x+\frac{1}{x})^2=\frac{(17+8)}{4}$

⇒$(x+\frac{1}{x})=\sqrt{\frac{25}{4}}$

⇒$(x+\frac{1}{x})=\frac{5}{2}$

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

⇒$(x^3+\frac{1}{x^3})=(\frac{5}{2})^3-3\times{(\frac{5}{2})}$

⇒$(x^3\frac{1}{x^3})=(\frac{125}{8})-(\frac{15}{2})$

⇒$(x^3+\frac{1}{x^3})=\frac{(125-60)}{8}$

⇒$(x^3+\frac{1}{x^3})=\frac{65}{8}\times(\frac{8}{13})$

⇒$\frac{8}{13}\times(x^3+\frac{1}{x^3})=5$

∴ The required value is 5.

Shortcut Trick

If $x^2+\frac{1}{x^2}=a$ , then $x+\frac{1}{x}=\sqrt{(a+2)}$

So,  if  $(x^2)^2+\frac{1}{(x^2)^2}=\frac{257}{16}$, then $x^2+\frac{1}{x^2}=\sqrt{\frac{257}{16}+2}=\frac{17}{4}$

Again, $x^2+\frac{1}{x^2}=\frac{17}{4}$ , then $x+\frac{1}{x}=\sqrt{\frac{17}{4}+2}=\frac{5}{2}$ 

Now, $(x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})$

⇒$\frac({5}{2})^3-3\times(\frac{5}{2})=(\frac{65}{8})$

∴  $(\frac{8}{13})\times(x^3+\frac{1}{x^3})=(\frac{65}{8})\times(\frac{8}{13})=5$