Answer
$\left\{-\dfrac{\sqrt[3]{175}}{5},1\right\}$
Work Step by Step
Using factoring of trinomials, the given equation, $
5x^6+2x^3-7=0
,$is equivalent to
\begin{align*}
(x^3-1)(5x^3+7)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l|r}
x^3-1=0 & 5x^3+7=0
\\
x^3=1 & 5x^3=-7
\\
& x^3=-\dfrac{7}{5}
.\end{array}
Taking the cube root of both sides and rationlizing the denominator (when necessary), the equations above are equivalent to
\begin{array}{l|r}
x=\sqrt[3]{1} & x=\sqrt[3]{-\dfrac{7}{5}}
\\\\
x=1 & x=\sqrt[3]{-\dfrac{7}{5}\cdot\dfrac{25}{25}}
\\\\
& x=\sqrt[3]{-\dfrac{1}{125}\cdot175}
\\\\
& x=\sqrt[3]{-\dfrac{1}{125}}\cdot\sqrt[3]{175}
\\\\
& x=-\dfrac{1}{5}\cdot\sqrt[3]{175}
\\\\
& x=-\dfrac{\sqrt[3]{175}}{5}
.\end{array}
Hence, the solution set of the equation $
5x^6+2x^3-7=0
$ is $\left\{-\dfrac{\sqrt[3]{175}}{5},1\right\}$.
