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 = [tex]\sqrt[3]{(15x-1)(13x+1)(x)}[/tex] .
Now, once again cubing on both sides, we get
(3x)³ = ([tex]\sqrt[3]{(15x-1)(13x+1)(x)}[/tex])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}
