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I am stuck in the following radical equation $$ (4x-1)\sqrt{x^3+1}=2x^3+2x+1. $$ SOLUTION I have tried based on the guidance of the commenters and I think that the following solution is rather simpler. $$ (4x-1)\sqrt{x^3+1}=2x^3+2x+1 $$ \begin{eqnarray*} &\Leftrightarrow& 2(x^3+1)-\sqrt{x^3+1}-4x\sqrt{x^3+1}+2x+2\sqrt{x^3+1}-1=0\\ &\Leftrightarrow& \sqrt{x^3+1}(2\sqrt{x^3+1}-1)-2x(2\sqrt{x^3+1}-1)+2\sqrt{x^3+1}-1=0\\ &\Leftrightarrow&(2\sqrt{x^3+1}-1)(\sqrt{x^3+1}-2x+2)=0\\ &\Leftrightarrow& 2\sqrt{x^3+1}-1=0\quad\text{or}\quad \sqrt{x^3+1}-2x+2=0\\ &\Leftrightarrow& x=\sqrt[3]{-3/4}\quad\text{or}\quad \begin{cases} x\geq 1/2\\ x^3+1=4x^2-4x+1\\ \end{cases}\\ &\Leftrightarrow& x=\sqrt[3]{-3/4}\quad\text{or}\quad x=2. \end{eqnarray*}

Blind
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    You’ll have to tell us what you’ve tried. Where do you get stuck? Is that the equation you started with or have you worked yourself into an algebra rut? – Dan Robertson Feb 01 '18 at 14:15

2 Answers2

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$$(4x-1)\sqrt{x^3+1}=2x^3+2x+1.$$ $$\left((4x-1)\sqrt{x^3+1}\right)^2=\left(2x^3+2x+1\right)^2$$ $$16x^5+16x^2-8x^4-8x+x^3+1=4x^6+8x^4+4x^3+4x^2+4x+1$$ $$4x^6-16x^5+16x^4+3x^3-12x^2+12x=0$$ $$x\left(x-2\right)^2\left(\sqrt[3]{4}x+\sqrt[3]{3}\right)\left(4^{\frac{2}{3}}x^2-\sqrt[3]{12}x+3^{\frac{2}{3}}\right)=0$$

$$\tag1 x-2=0 \implies x=2$$ $$\tag2\sqrt[3]{4}x+\sqrt[3]{3}=0 \implies x=-\sqrt[3]{\frac{3}{4}}$$ $$\tag3 4^{\frac{2}{3}}x^2-\sqrt[3]{12}x+3^{\frac{2}{3}}=0 \implies \nexists x\in \mathbb{R}$$ Therefore

$$x_1=2,\quad x_2=-\sqrt[3]{\frac{3}{4}}$$

Darío A. Gutiérrez
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after squaring and factorizing we obtain $$-x \left( 4\,{x}^{3}+3 \right) \left( x-2 \right) ^{2}=0$$ Can you solve this equation?

Dr. Sonnhard Graubner
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