Calculate the sum of all irrational roots of $4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$

Question

Calculate the sum of all irrational roots of $$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$

I'm not even sure how to begin here, I tried raising it to the power of three, tried writing $8x^{3}+ 3$ with $x^{3}+ y^{3}= \left ( x+ y \right )\left ( x^{2}- xy+ y^{2} \right ),$ but have had no meaningful results.

Answer 1

• If we assume "cube root" is the real-valued root, then we have

$$\begin{cases}4\sqrt[3]{8x-3}-8x^3-3=0 \\ x\in\mathbb R \end{cases} \Longleftrightarrow 512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0$$

$$\Longleftrightarrow (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0$$

Then using standard calculus tools (but, only algebraic ways are also possible ) we get

$$64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73 > 0$$

for all $x\in\mathbb R$.

So, we have only $2$ irrational roots:

$$x_1+x_2=\frac{-2}{4}=- \frac 12$$

Then, answer is equal to $- \dfrac 12.$

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• If we use the principal root instead of "real-valued cube root", then we have

$$4\sqrt[3]{8x-3}-8x^3-3=0 \Longleftrightarrow \begin{cases} 512 x^9 + 576 x^6 + 216 x^3 - 512 x + 219=0 \\ 8x-3≥0 \end{cases} $$

$$\Longleftrightarrow \begin{cases} (2 x - 1) (4 x^2 + 2 x - 3) (64 x^6 + 64 x^4 + 48 x^3 + 64 x^2 + 24 x + 73)=0 \\ x≥\frac 38\end{cases}$$

The last result implies, we have only $1$ irrational root:

$$x=\frac {-1+\sqrt {13}}{4}$$

So, answer is equal to $\dfrac {\sqrt {13}-1}{4}.$

Answer 2

Problem. Half-solve $$4\sqrt[3]{8x- 3}- 8x^{3}- 3= 0$$

Substitute $y^{3}= 8x- 3$ in the original equation_ $$\left ( 4y- 8x^{3}- 3 \right )+ \left ( y^{3}- 8x+ 3 \right )= \left ( y- 2x \right )\left ( 4x^{2}+ 2xy+ y^{2}+ 4 \right )= 0$$ With $y= 2x,$ we have the equivalent one $$8x^{3}- x+ 3=\!\left ( 2x- 1 \right )\!\left ( 4x^{2}+ 2x- 3 \right )=\!0\!\Rightarrow\exists_{x_{1}, x_{2}\in\left ( {\mathbb{Q}}' \right )^{2}}x_{1}+ x_{2}=\!-\frac{1}{2}$$ With $4x^{2}+ 2xy+ y^{2}+ 4= 0,$ there are no irrational roots, because $$4x^{2}+ 2xy+ y^{2}+ 4= y^{6}+ 4y^{4}+ 6y^{3}+ 16y^{2}+ 12y+ 73= 0\;{\rm illegal}$$