Given that $x\ne 0$ and $x$ is real that satisfies $\large \sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$. Find the product of all possible values of $x$.
Answer:
Let $\large a=\sqrt[3]{x^5-20x}=\sqrt[5]{x^3+20x}$.
Consider $\large a=\sqrt[3]{x^5-20x}$, we have:
$\large a^3=x^5-20x---(1)$
Consider $\large a=\sqrt[5]{x^3+20x}$, we have:
Showing posts with label find the product of all possible values of x. Show all posts
Showing posts with label find the product of all possible values of x. Show all posts
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