+0

# not a question, i just noticed is all. its about cube roots

+5
1123
6
+272

$${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{x}}}} = {\sqrt{{\frac{{\mathtt{x}}}{{\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{x}}}}}}}}$$

it solves out fine manually and all its just interesting.

Apr 26, 2015

#4
+112823
+10

Lets just proove this TheJonyMyster

$$\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\ RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\ RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\ RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\ RHS=x^{(1/n)}\\\\ RHS=\sqrt[n]{x}\\\\ RHS=LHS$$

Apr 28, 2015

#1
+32013
+10

It's true as long as x is greater than zero.

.

Apr 26, 2015
#2
+112823
+5

Yes TheJonyMyster, it is cool how roots work :)

Apr 26, 2015
#3
+272
+5

yeah, positive x cause of imaginary numbers. also it works for any $${\sqrt[{{\mathtt{{\mathtt{n}}}}}]{{\mathtt{x}}}} = {\sqrt[{{\mathtt{\left({\mathtt{n}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}}}]{{\frac{{\mathtt{x}}}{{\sqrt[{{\mathtt{{\mathtt{n}}}}}]{{\mathtt{x}}}}}}}}$$

Apr 26, 2015
#4
+112823
+10

Lets just proove this TheJonyMyster

$$\\\sqrt[n]{x}=\sqrt[(n-1)]{\frac{x}{\sqrt[n]{x}}}\\\\ RHS=\sqrt[(n-1)]{x^{1-\frac{1}{n}}}\\\\ RHS=\sqrt[(n-1)]{x^{\frac{n-1}{n}}}\\\\ RHS=x^{\frac{n-1}{n}}\right)^{\frac{1}{n-1}}\\\\ RHS=x^{(1/n)}\\\\ RHS=\sqrt[n]{x}\\\\ RHS=LHS$$

Melody Apr 28, 2015
#5
+117479
0

Nice proof,  Melody......!!!!

Apr 28, 2015
#6
+112823
0

thanks Chris :)

Apr 28, 2015