Question

Find all solutions to
$\sqrt[3]{15x-1} + \sqrt[3]{13x+1} =4\sqrt[3]{x}$
Enter all the separated, seperated by commas.

Answer 1

Answer:

x= 0 , $\frac{1}{14}$ , $\frac{-1}{12}$

Step-by-step explanation:

Given, equation is $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ = $4\sqrt[3]{x}$. →→→ (1)

Now, by cubing the equation on both sides, we get

( $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ )³ = ($4\sqrt[3]{x}$)³

⇒ (15x-1) + (13x+1) + 3×$\sqrt[3]{15x-1}$×$\sqrt[3]{13x+1}$ ($\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$) = 64 x.

⇒ 28x + 3×$\sqrt[3]{15x-1}$×$\sqrt[3]{13x+1}$ ($4\sqrt[3]{x}$) = 64x.        

(since from (1),  $\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ = $4\sqrt[3]{x}$)

⇒ 12× $\sqrt[3]{15x-1}$×$\sqrt[3]{13x+1}$ ($\sqrt[3]{x}$)= 36x.

⇒ 3x = $\sqrt[3]{(15x-1)(13+1)(x)}$.

Now, once again cubing on both sides, we get

(3x)³ = ($\sqrt[3]{(15x-1)(13+1)(x)}$)³.

⇒ 27x³ = (15x-1)(13x+1)(x).

⇒ 27x³ = 195x³ + 2x² - x

⇒ 168x³ + 2x² - x = 0

⇒ x(168x² + 2x -1) = 0

⇒ by, solving the equation we get ,

x = 0 ; x = $\frac{1}{14}$ ; x = $\frac{-1}{12}$

therefore, solution is x= 0 , $\frac{1}{14}$ , $\frac{-1}{12}$