# Terribly need help trying to simplify this equation

• Nov 5th 2010, 07:44 PM
CVB3
Terribly need help trying to simplify this equation
Hi guys! first post here and i have been trying to figure this one out for quite some time no and i really cant seem to crack it no matter where i look at it, which is why i turn to you guys, the only people who can help me. Actually i just dont follow the sequence as the answer was already shown, i just dont understand the transition:

= x^3 + y^3 + x^2y + xy^2

= (x+y)^3 - 3xy(x+y) + xy(x+y)

I really need you guys' help to understand this one, ive been at this for at least a good hour and still no luck. please help!
• Nov 5th 2010, 08:06 PM
pickslides
\$\displaystyle x^3 + y^3 = (x+y)(x^2-xy+y^2)\$

and

\$\displaystyle x^2y + xy^2 = xy(x+y)\$
• Nov 5th 2010, 08:21 PM
CVB3
Quote:

Originally Posted by pickslides
\$\displaystyle x^3 + y^3 = (x+y)(x^2-xy+y^2)\$

and

\$\displaystyle x^2y + xy^2 = xy(x+y)\$

thank you, thank you so much for the reply. I truly appreciate it, however, i still do not follow. I'm sorry, i'm not really the brightest person, i'm only in grade 8 but i really, really want to understand this and the fact that ive been doing this for 2 hours now and still cannot understand it really bothers me. Would it be possible if you could explain this to me?

I mean, when you moved \$\displaystyle x^3 + y^3\$ to the other side of the equation, should they become negative? and where did \$\displaystyle (x+y)(x^2-xy+y^2)\$ come from?

I can barely make out of the second line either, but im assuming that once i understand that first line in your post i would be able to work my way to ur second line.

Again, sorry for troubling you, and thank you for the time and effort, very much appreciated.
• Nov 5th 2010, 08:29 PM
pickslides
Quote:

Originally Posted by CVB3

I mean, when you moved \$\displaystyle x^3 + y^3\$ to the other side of the equation, should they become negative? and where did \$\displaystyle (x+y)(x^2-xy+y^2)\$ come from?

This is the rule for the sum of cubes. I am suggesting you factor your equation bit by bit.

For further info read here. 4. The Sum and Difference of Cubes

Quote:

Originally Posted by CVB3
I can barely make out of the second line either, but im assuming that once i understand that first line in your post i would be able to work my way to ur second line.

In the second line I have taken out \$\displaystyle xy\$ as a common factor in \$\displaystyle x^2y + xy^2\$ leaving \$\displaystyle xy(x+y)\$. This is the most fundamental idea in factorisation. If you don't understand this simple step there is no way you will be able to finish the entire problem. (Thinking)
• Nov 5th 2010, 08:34 PM
CVB3
Quote:

Originally Posted by pickslides
This is the rule for the sum of cubes. I am suggesting you factor your equation bit by bit.

For further info read here. 4. The Sum and Difference of Cubes

In the second line I have taken out \$\displaystyle xy\$ as a common factor in \$\displaystyle x^2y + xy^2\$ leaving \$\displaystyle xy(x+y)\$. This is the most fundamental idea in factorisation. If you don't understand this simple step there is no way you will be able to finish the entire problem. (Thinking)

thank you so much! i will read the link you have given me! like i said, thank you! i really came to the right place!