How do you solve $\sqrt[3]{7 n - 8} = 3$ $root3(7n-8)=3$ and find any extraneous solutions?

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How do you solve $\sqrt[3]{7 n - 8} = 3$ $root3(7n-8)=3$ and find any extraneous solutions?

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Answer 1 · 2016-08-18T13:36:41

$n = 5$ $n=5$ and there are no extraneous solutions.

Explanation:

$\sqrt[3]{7 n - 8} = 3$ $root3(7n-8)=3$

Taking cube of both sides we get
${(\sqrt[3]{7 n - 8})}^{3} = 3^{3}$ $(root3(7n-8))^3=3^3$
$\Rightarrow 7 n - 8 = 27$ $=>7n-8=27$

Rearranging and solving for $n$ $n$
$7 n = 27 + 8$ $7n=27+8$
$n = \frac{35}{7} = 5$ $n=35/7=5$
Insert $n = 5$ $n=5$ in original equation. We see that the equation holds good. We have a solution and there are no extraneous solutions.

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How do you solve $\sqrt[3]{7 n - 8} = 3$ $root3(7n-8)=3$ and find any extraneous solutions?

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How do you solve $\sqrt[3]{7 n - 8} = 3$ $root3(7n-8)=3$ and find any extraneous solutions?