Question

Find all solutions to

Answer 1

Answer:

Solution is x= 0 , x = \frac{1}{14}  , x = \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.

(since (a+b)³ = a³ + b³ + 3ab(a+b) ).

⇒ 28x + 3×\sqrt[3]{15x-1}× \sqrt[3]{13x+1} (\sqrt[3]{15x-1} +  \sqrt[3]{13x+1})  = 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]{15x-1}×\sqrt[3]{x}= 36x.

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

Now, once again cubing on both sides, we get

(3x)³ = ($\sqrt[3]{(15x-1)(13x+1)(x)}$)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 , x = \frac{1}{14}   ,  x = \frac{-1}{12}