I don't understand too well how this works

$\displaystyle \sqrt[3]{x+1=7}$

If I did that wrong please correct me . I saw it on my test today and did not understand it.

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- Oct 19th 2011, 04:41 PMvaironxxrdSolving for x in radicals
I don't understand too well how this works

$\displaystyle \sqrt[3]{x+1=7}$

If I did that wrong please correct me . I saw it on my test today and did not understand it. - Oct 19th 2011, 04:48 PMskeeterRe: Solving for x in radicals
- Oct 19th 2011, 04:57 PMvaironxxrdRe: Solving for x in radicals
- Oct 19th 2011, 05:03 PMskeeterRe: Solving for x in radicals
- Oct 19th 2011, 05:17 PMvaironxxrdRe: Solving for x in radicals
- Oct 19th 2011, 05:27 PMskeeterRe: Solving for x in radicals
that's all there is to it.

- Oct 19th 2011, 06:32 PMvaironxxrdRe: Solving for x in radicals
- Oct 19th 2011, 06:35 PMpsolakiRe: Solving for x in radicals
The question now is why should not have in mathematics formulas like:

$\displaystyle \sqrt[3]{x+1=7}$ and then taking to the cube the whole equation get an x=6??

Or even better solve 1st for x under the sqrt and then take 1/3 of 6 ?? - Oct 19th 2011, 06:42 PMWilmerRe: Solving for x in radicals
- Oct 19th 2011, 07:01 PMvaironxxrdRe: Solving for x in radicals
- Oct 20th 2011, 05:31 AMHallsofIvyRe: Solving for x in radicals
Because functions, such as the square root, are defined on

**numbers**, not equations. We sometimes talk about "taking the cube root" (or whatever other function you like) but what we mean is "take the cube root of each side".

You**can**for example, start with the equation, x+ 1= 7 and then take the cube root of both sides but that gives $\displaystyle \sqrt[3]{x+1}= \sqrt[3]{7}$, not $\displaystyle \sqrt[3]{x+ 1= 7}$. - Oct 20th 2011, 05:47 AMWilmerRe: Solving for x in radicals
- Oct 20th 2011, 06:05 AMYouklaRe: Solving for x in radicals