Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$

Question

Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$

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The root of equation very bad

My try 1:

Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$

Have: $a+b=4-x$

=>Root of a system of equation bad, too

My try 2:

Use quality $\sqrt[3]{a}\pm \sqrt[3]{b}=\frac{a\pm b}{\sqrt[3]{a^2}\mp \sqrt[3]{ab}+\sqrt[3]{b^2}}$ :

$\sqrt[3]{x+2}-ax+\sqrt[3]{2x-1}-bx=4-x$

Need find $ax$, $bx$ but it's very bad, too

Answer 1

to solve this equation use $$(a+b)^3=a^3+b^3+3ab(a+b)$$ $$(\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x)^3\\x+2+2x-1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3$$ put $\sqrt[3]{x+2}+\sqrt[3]{2x-1}=4-x \\$ so $$3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(\sqrt[3]{x+2}+\sqrt[3]{2x-1})=(4-x)^3\\ 3x+1+3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(4-x)=(4-x)^3\\3\sqrt[3]{x+2}.\sqrt[3]{2x-1}(4-x)=(4-x)^3-3x-1\\ 27(x+2)(2x-1)(4-x)^3=((4-x)^3-3x-1)^3$$ now you must go to the power of 3 and solve $ax^9+....=0$ degree=9
but it will be polynomial .

if you solve the eqaution with numerical or graphical method ,it has a root $x=1.325$ (the only root )

are you sure that type the equation correct ?

***why there is one real root ? if you take $f(x)=\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x-4 $ so $$f'(x)=\dfrac{1}{3\sqrt[3]{(x+2)^2}}+\dfrac{2}{3\sqrt[3]{(2x-1)^2}}+1 >0$$ so f(x) is strictly increasing $\to $ so $$f(x)=0$$ has only one root , It can be checked by $$f(1)<0 ,f(1.5)>0 \to x_0 \in (1,1.5)$$ so you can go on

split $(1,1.5) \to (1,1.25) ,(1.25,1.5)$ $$f(1)f(1.25)>0 \\f(1.25)f(1.5)<0 \to x_0 \in (1.25,1.5)$$ and take over

Answer 2

(Too long for a comment.)  Building upon this...

Let $\,\sqrt[3]{x+2}=a\,;\;\sqrt[3]{2x-1}=b$

The canonical way to get the polynomial equation in $x$ is to eliminate $a,b$ between:

$$ \begin{cases} \begin{align} a^3 &= x+2 \\ b^3 &= 2x-1 \\ a+b +x &= 4 \end{align} \end{cases} $$

This is a routine calculation using polynomial resultants, though generally not pretty to do by hand. In this case resultant[ resultant[ a+b+x-4, b^3-2x+1, b ], a^3-x-2, a ] gives $x$ as the root of:

$$ x^9 - 36 x^8 + 585 x^7 - 5589 x^6 + 34317 x^5 - 139563 x^4 + 374328 x^3 - 635553 x^2 + 615033 x - 253503 = 0 $$

Numerically, $x \;\;\simeq\;\; 1.3254785\dots$

Answer 3

If we set $a = \sqrt[3]{x+2}$ and $b = \sqrt[3]{2x-1}$, since $a + b = 4 - x = 1 + (3 - x)$, we obtain that

\begin{cases} a^{3} - b^{3} = a + b - 1\\ b^{3} = 2a^{3} - 5 \end{cases}

Hence we conclude that $b = 6 - a - a^{3}$, from whence we have the equation \begin{align*} (6 - a - a^{3})^{3} = 2a^{3} - 5 \end{align*} Which can be solved numerically. Wolfram, for example, gives $a \approx 1.49263$