Help with Simple Math Problems

[Maxwell]

Tonight my friend came to me with some simple math problems. Most of them were easy, but some of them I could not do, or could not remember how to do. It is very frustrating because I should know how to do this since I did it years ago.

I was wondering if you guys could help. The two problems I can't get are:

1) [tex] xy{\sqrt{125x^3y^5}} - {\sqrt{12x^5y^7}} + x^2y^3{\sqrt{{80xy}} [/tex]

2) [tex] {\sqrt[3]{16x^4y^3}} - 5x{\sqrt[3]{54xy^3}}} + 20y{\sqrt[3]{2x^4}} [/tex]

It seems so easy, jjust adding and subtracting, but I don't remember how to do this type of stuff.

Thanks in advance guys.

 

[cookiemonster]

[tex] xy{\sqrt{125x^3y^5}} - {\sqrt{12x^5y^7}} + x^2y^3{\sqrt{{80xy}} [/tex]

Let's do the first term.

[tex]xy{\sqrt{125x^3y^5}}[/tex]

Any powers of 2 can be brought out, so let's write it in terms of powers of 2:

[tex]xy\sqrt{(5^2)5(x^2)x(y^2)^2y}[/tex]

Bring out the squares

[tex]xy(5xy^2)\sqrt{5xy}[/tex]

Simplify

[tex]5x^2y^3\sqrt{5xy}[/tex]

Rinse and repeat for the remaining terms.

And for the cube root problem, powers of 3 can come out. So group them into powers of 3.

cookiemonster

 

[Maxwell]

Thank you Cookiemonster.

I still can't get it though.

 

[HallsofIvy]

[tex] {\sqrt[3]{16x^4y^3}} - 5x{\sqrt[3]{54xy^3}}} + 20y{\sqrt[3]{2x^4}} [/tex]

16= 8*2 and 8= 2³
x⁴= x*x³

so the first term is [itex]2xy\sqrt[3]{2x}[/itex]

54= 2*27= 2*3³
so the second term is [itex]5x(3y)\sqrt[3]{2x}[/itex]

In the third term x⁴= x*x³
so the third term is [itex]20y(x)\sqrt[3]{2x}[/itex]

That is: [itex]2xy\sqrt[3]{2x}+5x(3y)\sqrt[3]{2x}+20y(x)\sqrt[3]{2x}[/itex]

Notice the [itex]\sqrt[3]{2x}[/itex] in each term?

This is the same as [itex](2xy+ 15xy+ 20xy)\sqrt[3]{2x}[/itex]