Question

If $x=a^{\frac{1}{3}}+a^{\frac{-1}{3}}$, then find the value of $x^3-3x=a+\dfrac{1}{a}$.

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Answer 1

Answer :-

Given :-

• $\sf x = a^{\frac{1}{3}} + a^{\frac{-1}{3}}$

To Prove :-

• $\sf x^3 - 3x = a + \dfrac{1}{a}$

Solution :-

$\implies\sf x^3 - 3x$

$\implies\sf (a^{\frac{1}{3}} + a^{\frac{-1}{3}})^3 - 3(a^{\frac{1}{3}} + a^{\frac{-1}{3}})$

• $\sf (a + b)^3 = a^3 + b^3 + 3ab ( a + b)$

$\implies\sf (a^\frac{1}{3})^3 + (a^\frac{-1}{3})^3 + 3 \times a^\frac{1}{3} \times a^\frac{-1}{3} (a^{\frac{1}{3}} + a^{\frac{-1}{3}}) - 3(a^{\frac{1}{3}} + a^{\frac{-1}{3}})$

$\implies\sf a^{\frac{1}{\not3} \times \not3} + a^{\frac{-1}{\not3}\times \not3}+ 3 \times a^{\frac{1}{3}} + a^\frac{(-1)}{3}(a^\frac{1}{3} + a^\frac{-1}{3}) - 3(a^{\frac{1}{3}} + a^{\frac{-1}{3}})$

$\implies\sf a + a^{-1} + 3 \times a^0 \times (a^{\frac{1}{3}} + a^{\frac{-1}{3}}) - 3(a^{\frac{1}{3}} + a^{\frac{-1}{3}})$

$\implies\sf a + a^{-1} + \cancel{3(a^{\frac{1}{3}} + a^{\frac{-1}{3}})} - \cancel{3(a^{\frac{1}{3}} + a^{\frac{-1}{3}})}$

$\implies\sf a + a^{-1}$

$\implies\sf a + \dfrac{1}{a}$

$\boxed{\sf x^3 - 3x = a + \dfrac{1}{a}}$

Hence proved.