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If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression

$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$

My Try :

As I can find the value of $\displaystyle x$, from the given equation but it will be tedious I think !.

$$\displaystyle x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$$

$$\displaystyle \Longrightarrow \ x^{2} +1/2 \ =\ \frac{7}{2\sqrt{5}}$$

$$\displaystyle \Longrightarrow \ x^2 \ =\ \ \frac{7-\sqrt{5}}{2\sqrt{5}}$$

Which is getting too much complicated to solve the expression by putting the value of $\displaystyle x$.

What could be the other way to solve the given expression?

Girish Kumar Chandora
  • 931

3 Answers3

1

Let $$ a = \displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}}. $$ Notice that the first term is the multiplicative inverse of the second term. Thus $$ a^3 = \displaystyle \frac{x+1}{x-1} + \frac{x-1}{x+1} + 3 \left(\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \right). $$ So $$ a^3 - 3 a = \displaystyle \frac{x+1}{x-1} + \frac{x-1}{x+1}. $$ You can calculate the right hand side and then solve for $a$.

Slugger
  • 5,556
  • I also did this approach, as the equation will be cubic in terms of $a$ , which would again another task to solve the cubic expression for such typical number – Girish Kumar Chandora Aug 18 '20 at 10:36
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Let me try it

we have putting $t=\tan\Big(\frac{x_1}{2}\Big)$ $$\cos(x_1)=\frac{1-t^2}{1+t^2}$$

It's the Weierstrass substitution

Putting $y=t^2$ we get :

$$\cos(x_1)=\frac{1-y}{1+y}$$

Now we put $x=y$ to get :

$$\Big(-\cos(x_1)\Big)^{\frac{1}{3}}+\Big(-\frac{1}{\cos(x_1)}\Big)^{\frac{1}{3}}=?$$

After that I have tried $\cos(3x)=4\cos^3(x)-3\cos(x)$.

Hope it helps (I think it's hard).

Miss and Mister cassoulet char
  • 3,511
0

Consider $A = (\frac{x+1}{x-1})^\frac{1}{3} + (\frac{x-1}{x+1})^\frac{1}{3}$ and $x = \tan u$. Since $x^4+x^2 = \frac{11}{5}$ then $\frac{\sin^2u}{(1-\sin^2u)^2}=\frac{11}{5}$ by solving a simple quadric equation we get that $\sin u = \pm \sqrt{\frac{27-7\sqrt5}{22}}$. Therefore $A = (\frac{\tan u+1}{\tan u-1})^\frac{1}{3} + (\frac{\tan u+1}{\tan u-1})^\frac{1}{3}$ so from the tangent formula for adding two arc we get: $-A = \tan^\frac{1}{3} {(u+\frac{\pi}{4})}+ \cot^\frac{1}{3} {(u+\frac{\pi}{4})}$. By rasing both sides to 3 we get:$$3A-A^3=\pm\frac{1}{\sin(u+\frac{\pi}{4})\times\sqrt{1-\sin^2{(u+\frac{\pi}{4})}}}$$On the other hand we know: $\sin(u+\frac{\pi}{4})=\pm\frac{\sqrt {225\sqrt 5-280}}{11}$. By simplifications and solving a simple polynomial equation of third degree we can easily get the final answer. I hope that it can help you. Thank you for your question.

Amirhossein Dolatkhah
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