1. ## Reducing Roots

Expression $\sqrt[3]{16x^7y^5}$

1. $\sqrt[3]{2^32x^3x^3xy^3y^2}$
-I understand simplyifying in this step. $\sqrt[3]{16} = 2^3*2$

2. $\sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}$
-Here is where im confused. How and why are you able to separate the $2^3*2$ from $\sqrt[3]{16}$ into two separate locations? At the begging of the expresion the $2^3$ is present, while the other 2 is all the way at the end of the expression with $\sqrt[3]{2xy^2}$.

Many thanks for clarifying!

2. Originally Posted by allyourbass2212
Expression $\sqrt[3]{16x^7y^5}$

1. $\sqrt[3]{2^32x^3x^3xy^3y^2}$
-I understand simplyifying in this step. $\sqrt[3]{16} = 2^3*2$

2. $\sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}$
-Here is where im confused. How and why are you able to separate the $2^3*2$ from $\sqrt[3]{16}$ into two separate locations? At the begging of the expresion the $2^3$ is present, while the other 2 is all the way at the end of the expression with $\sqrt[3]{2xy^2}$.

Many thanks for clarifying!
look
you know that

$\sqrt[3]{ab}=(\sqrt[3]{a})(\sqrt[3]{b})$

$2(4)=\sqrt{4}(4)=\sqrt{4}(\sqrt{16})=\sqrt{4(16)}= \sqrt{64}=8$

$\sqrt[3]{(2)(2)(2)}=2$

the square degree is 3 so I take one thing from every three thing

$\sqrt[3]{}=\sqrt[3]{}" alt="\sqrt[3]{}=\sqrt[3]{}" />

$\sqrt[4]{{\color{red}2*2*2*2}*2*{\color{blue}3*3*3*3}}=2(3 )\sqrt[4]{2}$

when I said

2^4=2(2)(2)(2) four times

take the fourth sqrt

$\sqrt[4]{(2)(2)(2)(2)}=2$

if

$\sqrt{2(2)(2)}=2\sqrt{2}$

when we said the square of degree four of A that's mean
we need a number if you multiply itself by itself four times we get A

it is clear or not

3. I understand your examples, but im sorry I do not grasp whats going on from here $\sqrt[3]{2^32x^3x^3xy^3y^2}$ to $\sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}$. Specifically regarding the movement of the 2 from the beginning to the end of the expression.

4. Originally Posted by allyourbass2212
I understand your examples, but im sorry I do not grasp whats going on from here $\sqrt[3]{2^32x^3x^3xy^3y^2}$ to $\sqrt[3]{2^3} \sqrt[3]{x^3} \sqrt[3]{x^3} \sqrt[3]{y^3} \sqrt[3]{2xy^2}$. Specifically regarding the movement of the 2x to the end of the expression.

you know that

abcd=acbd=bacd=dacb

$\sqrt[3]{16x^7y^5}$

lats simplify it
you know that y^5=(y)(y)(y)(y)(y)
x^7=x(x)(x)(x)(x)(x)(x)
16=2(2)(2)(2)
for each three same thing take one

$\sqrt[3]{16y^5x^7}=\sqrt[3]{2{\color{red}(2)(2)(2)}(y)(y){\color{blue}(7)(7)( 7)}x{\color{green}(x)(x)(x)}{\color{red}(x)(x)(x)} }$
you will take two x since you have 6 x's and one y since you have three y and one 2 since you have three what will stay in the sqrt

$(x)(x)(y)(2)\sqrt[3]{2(y)(y)(x)}$

now collect them
like this
$(x)(x)=x^2$

$2x^2(y)\sqrt[3]{2y^2x}$

you can change the order since the multiple is a commutative operation

$2(y)x^2\sqrt[3]{2xy^2}$

I think it is clear

5. Thanks Amer, that method makes sense!

I am not sure what the book is trying to do but thats not important at the moment.

The only question I have is how do you know $\sqrt[3]{16}$=
(2)(2)(2)(2)

And could you list the break down of some other numbers for example?

Thanks!

6. Originally Posted by allyourbass2212
Thanks Amer, that method makes sense!

I am not sure what the book is trying to do but thats not important at the moment.

The only question I have is how do you know $\sqrt[3]{16}$=
(2)(2)(2)(2)

And could you list the break down of some other numbers for example?

Thanks!
ok sure

$\sqrt[3]{16}=$ $\sqrt[3]{2{\color{red}(2)(2)(2)}}=2\sqrt[3]{2}$

easy one

$\sqrt{144}=\sqrt{2(72)}=\sqrt{{\color{red}2(2)}(36 )}=2\sqrt{36}=2\sqrt{2(2)(3)(3)}=2(2)(3)=12$

level is medium

$\sqrt[3]{864}=\sqrt[3]{2(432)}=\sqrt[3]{2(2)(216)};=\sqrt[3]{{\color{red}(2)(2)(2)}(108)}=2\sqrt[3]{108}=$
$2\sqrt[3]{2(54)}=2\sqrt[3]{2(2)(27)}=2\sqrt[3]{(2)(2)(3)(9)}=2\sqrt[3]{(2)(2){\color{blue}(3)(3)(3)}}=2(3)\sqrt[3]{2^2}$

hard one here I will not expand like before I will write $a(a)(a)=a^3$ right

$\sqrt[4]{y^15x^5z^7}=\sqrt[4]{{\color{blue}(y^4)(y^4)(y^4)}(y^3){\color{red}(x^ 4)}(x){\color{red}(z^4)}(z^3)}$ $=y(y)(y)(x)(z)\sqrt[4]{y^3(x)(z^3)}=y^3(x)(z)\sqrt[4]{y^3(x)(z^3)}$

if there is any question feel free to post it .........