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Question: If \( x + \dfrac{1}{x} = 2\cos \theta \) ,then \( {x^3} + \dfrac{1}{{{x^3}}} \) is equal to...

If x+1x=2cos⁡θx + \dfrac{1}{x} = 2\cos \theta ,then x3+1x3{x^3} + \dfrac{1}{{{x^3}}} is equal to

Explanation

Solution

Hint : ake cube of LHS & RHS and then use trigonometric identities
Initially rewrite the given equation LHS part so that you will have an idea what to solve.After that try cubing on both sides because the RHS part is in the power of the cube hence first try cubing so that the equation will expand and you will have more room to solve it. After that expand both the sides according to the formulas. Then start substitute the first LSH value into the present equation so that the RHS will have more terms. After that convert the variable form of the equation to trigonometric form so that the RHS will be more simplified to the shorter form. Hence by following all these steps we will have the required solution.

Complete step-by-step answer :
Given the question consists of,
x+1x=2cos⁡θx + \dfrac{1}{x} = 2\cos \theta
Now, take cube of both LHS and RHS
This step will give us
(x+1x)3=(2cos⁡θ)3{\left( {x + \dfrac{1}{x}} \right)^3} = {(2\cos \theta )^3}
Now, use (a+b)3{(a + b)^3} formula to the LHS which equals to a3+b3+3a2b+3ab2{a^3} + {b^3} + 3{a^2}b + 3a{b^2}
So, our new LHS becomes
x3+1x3+3x×1x(x+1x)=8cos⁡3θ{x^3} + \dfrac{1}{{{x^3}}} + 3x \times \dfrac{1}{x}\left( {x + \dfrac{1}{x}} \right) = 8{\cos ^3}\theta
Solving this further we will get

x3+1x3+3x×1x(x+1x)=8cos⁡2θ ⇒x3+1x3+3(x+1x)=8cos⁡2θ   {x^3} + \dfrac{1}{{{x^3}}} + 3x \times \dfrac{1}{x}\left( {x + \dfrac{1}{x}} \right) = 8{\cos ^2}\theta \\\ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x + \dfrac{1}{x}} \right) = 8{\cos ^2}\theta \;

But we know that x+1x=2cos⁡θx + \dfrac{1}{x} = 2\cos \theta
Replacing this in the equation we will get

x3+1x3+3(x+1x)=8cos⁡2θ ⇒x3+1x3+3(2cos⁡θ)=8cos⁡3θ ⇒x3+1x3+6cos⁡θ=8cos⁡3θ   {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x + \dfrac{1}{x}} \right) = 8{\cos ^2}\theta \\\ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {2\cos \theta } \right) = 8{\cos ^3}\theta \\\ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 6\cos \theta = 8{\cos ^3}\theta \;

Taking the trigonometric terms on the RHS, we will get

x3+1x3+6cos⁡θ=8cos⁡3θ ⇒x3+1x3=8cos⁡3θ−6cos⁡θ ⇒x3+1x3=2(4cos⁡2θ−3cos⁡θ); {x^3} + \dfrac{1}{{{x^3}}} + 6\cos \theta = 8{\cos ^3}\theta \\\ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 8{\cos ^3}\theta - 6\cos \theta \\\ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 2\left( {4{{\cos }^2}\theta - 3\cos \theta } \right) ;

Now, terms inside the brackets is a direct formula of cos⁡3θ\cos 3\theta
Therefore, replacing it we will get

⇒x3+1x3=8cos⁡3θ−6cos⁡θ ⇒x3+1x3=2cos3θ   \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 8{\cos ^3}\theta - 6\cos \theta \\\ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 2cos3\theta \;

Hence, we have our required solution as stated in the question is 2cos3θ2cos3\theta .
So, the correct answer is “ 2cos3θ2cos3\theta ”.

Note : As mentioned earlier trigonometric simplification may be tough but by using some simple method like converting them into shorter forms it becomes easy to solve them. We have to make sure while placing the correct value after squaring or cubing of the expression or the question or else the result gets completely varied.