Math Doubts

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.



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