Question

if x =3^1/3 + 3^-1/3 then find the value of 3x^3-9x
​

Answer 1

Answer:

Let's solve each term on the RHS of the first equation:

3

1

3

=

3

√

3

3

−

1

3

=

1

3

1

3

=

1

3

√

3

∴

x

=

(

3

√

3

+

1

3

√

3

)

Plugging in that value for

x

yields:

3

(

3

√

3

+

1

3

√

3

)

3

−

9

(

3

√

3

+

1

3

√

3

)

=

10

That's the extent of my knowledge. I'll have someone check and add on to this answer.

Answer 2

Step-by-step explanation:

$\sf{Given : \; x = 3^{\frac{1}{3}} + 3^{\frac{-1}{3}}}\\\\\\\textsf{Cubing on both sides, We get :}\\\\\\\implies x^3 = (3^{\frac{1}{3}} + 3^{\frac{-1}{3}})^3\\\\\\\bigstar\;\;We\;know\;that : \boxed{\sf{(a + b)^3 = a^3 + b^3 + 3ab^2 + 3a^2b}}$

$\sf{\implies x^3 = (3^\frac{1}{3})^3 + (3^\frac{-1}{3})^3 + 3(3^\frac{1}{3})(3^\frac{-1}{3})^2 + 3(3^\frac{1}{3})^2(3^\frac{-1}{3})}$

$\sf{\bigstar\;\;We\;know\;that : \boxed{\sf{(a^\frac{1}{n})^m = a^\frac{m}{n}}}}$

$\sf{\implies x^3 = (3^\frac{3}{3}) + (3^\frac{-3}{3}) + 3(3^\frac{1}{3})(3^\frac{-2}{3}) + 3(3^\frac{2}{3})(3^\frac{-1}{3})}$

$\sf{\bigstar\;\;We\;know\;that : \boxed{\sf{(a)^m \times (a)^n = a^{m + n}}}}$

$\sf{\implies x^3 = 3 + 3^{-1} + 3^{\big(1 + \frac{1}{3} - \frac{2}{3}\big)} + 3^{\big(1 - \frac{1}{3} + \frac{2}{3}\big)}}$

$\sf{\implies x^3 = 3 + \dfrac{1}{3} + 3^{\big(1 - \frac{1}{3}\big)} + 3^{\big(1 + \frac{1}{3}\big)}}$

$\sf{\implies x^3 = 3 + \dfrac{1}{3} + 3^{(\frac{2}{3})} + 3^{(\frac{4}{3})}}$

$\textsf{Multiplying the above Equation with 3, We get :}$

$\sf{\implies 3x^3 = (3 \times 3) + \dfrac{3}{3} + 3^{(\frac{2}{3})}(3) + 3^{(\frac{4}{3})}(3)}$

$\sf{\implies 3x^3 = 9 + 1 + 3^{(\frac{2}{3} + 1)} + 3^{(\frac{4}{3} + 1)}}$

$\sf{\implies 3x^3 = 10 + 3^{(\frac{5}{3})} + 3^{(\frac{7}{3})}}$

$\sf{\implies 3x^3 = 10 + 3^{(\frac{5}{3})} + 3^{(\frac{7}{3})}\;-----\;[1]}$

$\sf{Now,\;Consider :\;x = 3^{\frac{1}{3}} + 3^{\frac{-1}{3}}}$

$\textsf{Multiplying the above Equation with 9, We get :}$

$\sf{\implies 9x = 3^{\frac{1}{3}}(9) + 3^{\frac{-1}{3}}(9)}$

$\sf{\implies 9x = 3^{\frac{1}{3}}(3^2) + 3^{\frac{-1}{3}}(3^2)}$

$\sf{\implies 9x = 3^{\big(\frac{1}{3} + 2\big)} + 3^{\big(\frac{-1}{3} + 2\big)}}$

$\sf{\implies 9x = 3^{(\frac{7}{3})} + 3^{(\frac{5}{3})}\;-----\;[2]}$

$\textsf{Now, Subtracting Equation [2] from Equation [1], We get :}$

$\sf{\implies 3x^3 - 9x = 10 + 3^{(\frac{5}{3})} + 3^{(\frac{7}{3})} - [3^{(\frac{5}{3})} + 3^{(\frac{7}{3})}]}$

$\sf{\implies 3x^3 - 9x = 10 + 3^{(\frac{5}{3})} + 3^{(\frac{7}{3})} - 3^{(\frac{5}{3})} - 3^{(\frac{7}{3})}}$

$\sf{\implies 3x^3 - 9x = 10}$