# Simple math problems

1. May 13, 2004

### 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 cant get are:

1) $$xy{\sqrt{125x^3y^5}} - {\sqrt{12x^5y^7}} + x^2y^3{\sqrt{{80xy}}$$

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

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

2. May 13, 2004

$$xy{\sqrt{125x^3y^5}} - {\sqrt{12x^5y^7}} + x^2y^3{\sqrt{{80xy}}$$

Let's do the first term.

$$xy{\sqrt{125x^3y^5}}$$

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

$$xy\sqrt{(5^2)5(x^2)x(y^2)^2y}$$

Bring out the squares

$$xy(5xy^2)\sqrt{5xy}$$

Simplify

$$5x^2y^3\sqrt{5xy}$$

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.

3. May 13, 2004

### Maxwell

I still can't get it though.

4. May 13, 2004

### HallsofIvy

Staff Emeritus
$${\sqrt[3]{16x^4y^3}} - 5x{\sqrt[3]{54xy^3}}} + 20y{\sqrt[3]{2x^4}}$$

16= 8*2 and 8= 23
x4= x*x3

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

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

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

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

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

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