What is $\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} ?$ $root(3)x-1/(root(3)x)?$

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What is $\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} ?$ $root(3)x-1/(root(3)x)?$

Please explain me how it says $\sqrt[3]{x} \sqrt[3]{x} = x^{\frac{2}{3}}$ $root(3)xroot(3)x=x^(2/3)$

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Answer 1 · 2016-03-02T14:08:18

$\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}$ $root(3)x-1/(root(3)x)$

Take out the $L C D : \sqrt[3]{x}$ $LCD:root(3)x$

$\to \frac{\sqrt[3]{x} \cdot \sqrt[3]{x}}{\sqrt[3]{x}} - \frac{1}{\sqrt[3]{x}}$ $rarr(root(3)x*root(3)x)/root(3)x-1/(root(3)x)$

Make their denominators same

$\to \frac{(\sqrt[3]{x} \cdot \sqrt[3]{x}) - 1}{\sqrt[3]{x}}$ $rarr((root(3)x*root(3)x)-1)/(root(3)x)$

$\sqrt[3]{x} \cdot \sqrt[3]{x} = \sqrt[3]{x \cdot x} = \sqrt[3]{x^{2}} = x^{\frac{2}{3}}$ $root(3)x*root(3)x=root(3)(x*x)=root(3)(x^2)=x^(2/3)$

$\Rightarrow = \frac{x^{\frac{2}{3}} - 1}{\sqrt[3]{x}}$ $rArr=(x^(2/3)-1)/root(3)(x)$

Answer 2 · 2016-03-02T15:41:32

$Explaining the connection between \sqrt[3]{x} \sqrt[3]{x} and x^{\frac{2}{3}}$ $color(blue)("Explaining the connection between "root(3)(x)root(3)(x)" and "x^(2/3))$

Explanation:

$Point 1$ $color(blue)("Point 1")$

Look at these alternative ways of writing roots

$\sqrt{x} is the same as x^{\frac{1}{2}}$ $sqrt(x)" is the same as " x^(1/2)$

$\sqrt[3]{x} is the same as x^{\frac{1}{3}}$ $root(3)(x)" is the same as "x^(1/3)$

$\sqrt[4]{x} is the same as x^{\frac{1}{4}}$ $root(4)(x)" is the same as "x^(1/4)$

So for any number $n \sqrt[n]{x} is the same as x^{\frac{1}{n}}$ $n" "root(n)(x) " is the same as "x^(1/n)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Point 2$ $color(blue)("Point 2")$

Just picking a number at random I chose 3

Another way (not normally done) of writing 3 is $3^{1}$ $3^1$

When you have $3 \times 3 it can be written as 3^{2}$ $3xx3" it can be written as " 3^2$

In the same way $3 \times 3 \times 3 can be written as 3^{3}$ $3xx3xx3" can be written as "3^3$

In the same way $3 \times 3 \times 3 \times 3 can be written as 3^{4}$ $3xx3xx3xx3" can be written as "3^4$

Notice that $3 \times 3 = 3^{1} \times 3^{1} = 3^{1 + 1} = 3^{2}$ $3xx3 = 3^1xx3^1 =3^(1+1) = 3^2$

Notice that $3 \times 3 \times 3 = 3^{1} \times 3^{1} \times 3^{1} = 3^{1 + 1 + 1} = 3^{3}$ $3xx3xx3=3^1xx3^1xx3^1=3^(1+1+1)=3^3$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Point 3$ $color(blue)("Point 3")$

Given that a way of writing square root of 3 is $\sqrt{3} is 3^{\frac{1}{2}}$ $sqrt(3)" is "3^(1/2)$

Compare what happens in each of the following two rows

$3^{1} \times 3^{1} \times 3^{1} = 3^{1 + 1 + 1} = 3^{3}$ $3^1xx3^1xx3^1 = 3^(1+1+1) = 3^3$

$3^{\frac{1}{2}} \times 3^{\frac{1}{2}} \times 3^{\frac{1}{2}} = 3^{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}} = 3^{\frac{3}{2}}$ $3^(1/2)xx3^(1/2)xx3^(1/2) =3^(1/2+1/2+1/2)=3^(3/2)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Point 4$ $color(blue)("Point 4")$

$You asked about \sqrt[3]{x} \sqrt[3]{x} = x^{\frac{2}{3}}$ $color(brown)("You asked about "root(3)(x)root(3)(x)=x^(2/3))$

From above we know that $\sqrt[3]{x} is the same as x^{\frac{1}{3}}$ $root(3)(x)" is the same as "x^(1/3)$

But we have $\sqrt[3]{x} \sqrt[3]{x}$ $root(3)(x)root(3)(x)$

This is the same as $x^{\frac{1}{3}} \times x^{\frac{1}{3}} = x^{\frac{1}{3} + \frac{1}{3}} = x^{\frac{2}{3}}$ $x^(1/3)xxx^(1/3) = x^(1/3+1/3) = x^(2/3)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Point 5$ $color(blue)("Point 5")$

Backtrack for a moment and again think about

$x^{\frac{1}{3}} \times x^{\frac{1}{3}}$ $x^(1/3)xxx^(1/3)$

Like in $3 \times 3 = 3^{2}$ $3xx3 = 3^2$

$x^{\frac{1}{3}} \times x^{\frac{1}{3}} = {(x^{\frac{1}{3}})}^{2}$ $x^(1/3)xxx^(1/3)=(x^(1/3))^2$

and $x^{\frac{1}{3}} \times x^{\frac{1}{3}} = x^{\frac{1}{3} + \frac{1}{3}} = x^{\frac{2}{3}}$ $x^(1/3)xxx^(1/3)=x^(1/3+1/3)=x^(2/3)$

Then ${(x^{\frac{1}{3}})}^{2} = x^{\frac{1 \times 2}{3}} = x^{\frac{2}{3}}$ $(x^((color(magenta)(1))/3))^(color(green)(2)) = x^((color(magenta)(1)xxcolor(green)(2))/3) = x^(2/3)$

Turning this back the other way

$x^{\frac{2}{3}} = \sqrt[3]{x^{2}}$ $x^(2/3) = root(3)(x^2)$

Practise and a lot of it will fix this in your mind. It will seem confusing at first but as you practise more and more it will suddenly click!

Hope this helps!!

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2 Answers

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