# Evaluate $x^3+\dfrac{1}{x^3}$ if $x+\dfrac{1}{x} = 2\cos{\theta}$

The sum of a variable and its reciprocal is equal to twice the cos of angle theta.

$x+\dfrac{1}{x} = 2\cos{\theta}$

Then, the value of sum of cube of variable and cube of its reciprocal is required to find.

### Take cube both sides

The cube of algebraic trigonometric equation should be evaluated to find the value of sum of cube of variable and cube of its reciprocal.

$\implies$ ${\Bigg(x+\dfrac{1}{x}\Bigg)}^3 = {({2\cos{\theta}})}^3$

### Use cube of sum of terms rule

The cube of sum of two terms can be expanded by the a plus b whole cube formula.

${(a+b)}^3 = a^3+b^3+3ab{(a+b)}$

$\implies$ $x^3+{\Bigg(\dfrac{1}{x}\Bigg)}^3 + 3 \times x \times \dfrac{1}{x} \Bigg(x+\dfrac{1}{x}\Bigg)$ $\,=\,$ $2^3\cos^3{\theta}$

$\implies$ $\require{cancel} x^3+\dfrac{1}{x^3} + 3\dfrac{\cancel{x}}{\cancel{x}} \Bigg(x+\dfrac{1}{x}\Bigg)$ $\,=\,$ $8\cos^3{\theta}$

### Simplifying the equation

$\implies$ $x^3+\dfrac{1}{x^3} + 3\Bigg(x+\dfrac{1}{x}\Bigg)$ $\,=\,$ $8\cos^3{\theta}$

It is given in this problem that the sum of a variable and its reciprocal is equal to twice the cosine of angle theta. So, replace it by $2\cos{\theta}$.

$\implies$ $x^3+\dfrac{1}{x^3} + 3{(2\cos{\theta})}$ $\,=\,$ $8\cos^3{\theta}$

$\implies$ $x^3+\dfrac{1}{x^3} + 6\cos{\theta}$ $\,=\,$ $8\cos^3{\theta}$

$\implies$ $x^3+\dfrac{1}{x^3}$ $\,=\,$ $8\cos^3{\theta}-6\cos{\theta}$

$\implies$ $x^3+\dfrac{1}{x^3}$ $\,=\,$ $2{(4\cos^3{\theta}-3\cos{\theta})}$

According to cos triple angle formula, the right hand side trigonometric expression can be written as cos of angle $3\theta$.

$\,\,\, \therefore \,\,\,\,\,\,$ $x^3+\dfrac{1}{x^3}$ $\,=\,$ $2\cos{3\theta}$

It is the required solution for this algebraic trigonometric math problem.