Cube roots equation

Relevant wiki: Radical Equations - Intermediate

Solution suggested by @Harsh Shrivastava

$\begin{aligned} \sqrt[3]{15x-1} + \sqrt[3]{13x+1} & = 4 \sqrt[3]{x} \\ \implies \sqrt[3]{15x-1} + \sqrt[3]{13x+1} + (-4 \sqrt[3]{x}) & = 0 \\ \implies (\sqrt[3]{15x-1})^3 + (\sqrt[3]{13x+1})^3 + (-4 \sqrt[3]{x})^3 & = 3(\sqrt[3]{15x-1})(\sqrt[3]{13x+1})(-4 \sqrt[3]{x}) \\ 15x-1 + 13x+1 - 64x & = -12 \sqrt[3]{195x^3 + 2x^2 - x} \\ -36x & = -12\sqrt[3]{195x^3 + 2x^2 - x} \\ 3x & = \sqrt[3]{195x^3 + 2x^2 - x} \\ 27x^3 & = 195x^3 + 2x^2 - x \\ \implies 168x^3+2x^2-x & = 0 \\ x(14x-1)(12x+1) & = 0 \\ \implies x & = 0, \frac 1{14}, -\frac 1{12} \end{aligned}$

Therefore, the sum of roots $= 0 +\frac 1{14} - \frac 1{12} = - \frac 1{84}$ $\implies a+b = 85$ .

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My solution

$\begin{aligned} \sqrt[3]{15x-1} + \sqrt[3]{13x+1} & = 4 \sqrt[3]{x} \quad \quad \small \color{#3D99F6}{\text{Cubing both sides}} \\ 15x-1 + 3(\sqrt[3]{15x-1})^2\sqrt[3]{13x+1} + 3\sqrt[3]{15x-1}(\sqrt[3]{13x+1})^2 + 13x + 1 & = 64 x \\ 3\sqrt[3]{(15x-1)(13x+1)} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1} ) & = 64x - 28x \\ 3\sqrt[3]{(15x-1)(13x+1)} (4 \sqrt[3]{x} ) & = 36x \\ \sqrt[3]{x(15x-1)(13x+1)} & = 3x \quad \quad \small \color{#3D99F6}{\text{Cubing both sides and expand}} \\ 195x^3 + 2x^2 - x & = 27x^3 \\ \implies 168x^3+2x^2-x & = 0 \\ x(14x-1)(12x+1) & = 0 \\ \implies x & = 0, \frac 1{14}, -\frac 1{12} \end{aligned}$

Therefore, the sum of roots $= 0 +\frac 1{14} - \frac 1{12} = - \frac 1{84}$ $\implies a+b = 85$ .

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

$\sqrt[3] {14x + (x - 1)} + \sqrt[3]{14x - (x - 1)} = 4\sqrt[3]{x}$

Cubing both sides we get:

${14x + (x - 1)} + {14x - (x - 1)} + 3\sqrt[3]{196x^{2} - (x - 1)^{2}} (4\sqrt[3]{x}) = 64x$

${28x + 12\sqrt[3]{x(195x^{2} + 2x - 1)} = 64x}$

${12\sqrt[3]{x(195x^{2} + 2x - 1)} = 36x}$

${\sqrt[3]{x(195x^{2} + 2x - 1)} = 3x}$

Cubing both sides again we get:

${{x(195x^{2} + 2x - 1)} = 27x^{3}}$

Since we have x as a common factor that means x = 0 is one of the solutions but since we are calculating the sum of all roots, we can just simplify by x all terms without affecting the answer.

${{195x^{2} + 2x - 1} = 27x^{2}}$

${{168x^{2} + 2x - 1} = 0}$

This equation has 2 real solutions since the discriminant is positive. The sum of the roots is:

$\frac{-2}{168} = \frac{-1}{84}$

$a+b = 85$

$\sqrt[3]{15x-1}$ + $\sqrt[3]{13x+1}$ =4 $\sqrt[3]{x}$
Cubing both sides we get

$15x-1$ + $13x+1$ + 3( $\sqrt[3]{15x-1})^2$ $\sqrt[3]{13x+1}$ + 3 $\sqrt[3]{15x-1}$ ( $\sqrt[3]{13x+1})^2$ = $64x$

3( $\sqrt[3]{15x-1})^2$ ( $\sqrt[3]{13x+1}$ ) + 3 $(\sqrt[3]{15x-1}$ ) ( $\sqrt[3]{13x+1})^2$ = $36x$

( $\sqrt[3]{15x-1})^2$ ( $\sqrt[3]{13x+1}$ ) + $(\sqrt[3]{15x-1}$ ) ( $\sqrt[3]{13x+1})^2$ = $12x$

Assume that $(\sqrt[3]{15x-1}$ ) = $A$ and ( $\sqrt[3]{13x+1}$ ) = $B$

We will get;

$A^2$ $B$ + $A$ $B^2$ = $\frac{3}{7}$ $(A^3$ + $B^3$ )

$A$ $B$ $(A$ + $B$ )= $\frac{3}{7}$ $(A$ + $B$ ) $(A^2$ - $A$ $B$ + $B^2$ )

$A$ $B$ = $\frac{3}{7}$ $(A^2$ - $A$ $B$ + $B^2$ )

Times 7 both sides

7 $A$ $B$ =3 $(A^2$ - $A$ $B$ + $B^2$ )

7 $A$ $B$ =3 $A^2$ -3 $A$ $B$ +3 $B^2$

3 $A^2$ -10 $A$ $B$ +3 $B^2$ =0

$(3A$ - $B$ ) $(A$ - $3B$ )=0

Therefore $B$ = $3A$ $A$ = $3B$

Substitute; $A$ and $B$ back

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

Cubing both sides

27( ${15x-1}$ )=( ${13x+1}$ )

${405x-27}$ = ${13x+1}$ Therefore;

${x}$ = $\frac{1}{14}$

and

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

Cubing both sides

( ${15x-1}$ )=27( ${13x+1}$ )

${15x-1}$ = ${351x+27}$ Therefore;

${x}$ = - $\frac{1}{12}$

Therefore; sum of the roots;

- $\frac{1}{12}$ + $\frac{1}{14}$ = - $\frac{1}{84}$

Therefore the answer is 85