Question

If X=3^1/3 + 3^-1/3,then 3x^3 - 9x is ?​

Answer 1

Answer:

18

Step-by-step explanation:

3^1/3 + 3^-1/3 =X

3^1-1/3 =X( taking lcm 3 and power are added if base is same)

so, 3^0/3 = X

therefore,X=1 (power 0 means nothing so 3/3 = 1)

now putting value of x

3x^3-9x=??

so  value is 18...

im my view so dnt know is it correct or not

Answer 2

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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