I was trying to solve this equation without using calculus. Is it possible to be solved by elementary algebraic methods?
$$\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$$
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where does the first root end? – Dr. Sonnhard Graubner Feb 01 '16 at 19:39
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1What happens if you raise both the third power, simplify, and square, simplify and solve. – fleablood Feb 01 '16 at 19:44
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yes it can solved by algebraic methods – Dr. Sonnhard Graubner Feb 01 '16 at 19:52
2 Answers
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Let $u=\sqrt[3]{q+\sqrt{p^{3}+q^{2}}}$, $v=\sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$,
we have $y=u+v$ satisfying $y^{3}+3py=2q \ldots \ldots (*)$.
Take $y=\sqrt[3]{2}$, $q=7x$ and $p^{3}+q^{2}=19^{2}$ or $p^{3}=19^{2}-(7x)^{2}$.
Substitute into $(*)$,
\begin{align*} 2+3\sqrt[3]{19^{2}-(7x)^{2}} \times \sqrt[3]{2} &= 2(7x) \\ \sqrt[3]{19^{2}-(7x)^{2}} &= \frac{14x-2}{3\sqrt[3]{2}} \\ 19^{2}-(7x)^{2} &= \frac{(14x-2)^{3}}{54} \end{align*}
By factor theorem or Cardano formula,
$\displaystyle x=\frac{7}{4}, \frac{-8\pm 3i\sqrt{15}}{7}$ subject to correct choices of cubic roots in the original equation.
P.S.: May rewrite as $\sqrt[3]{19+7x}-\sqrt[3]{19-7x}=\sqrt[3]{2}$ for easy computer verification.
Ng Chung Tak
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powering by $3$ we get $$7x+19+3\sqrt[3]{7x+19}^2\sqrt[3]{7x-19}+3\sqrt{7x+19}(\sqrt[3]{7x-19})^2+7x-19=2$$ and we get $$14x+3\sqrt[3]{7x+19}\sqrt[3]{7x-19}(\sqrt[3]{7x+19}+\sqrt[3]{7x-19})=2$$ $$14x+3\sqrt[3]{7x+19}\sqrt[3]{7x-19}\sqrt[3]{2}=2$$ can you proceed?
Dr. Sonnhard Graubner
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