SOLUTION: Im sorry if the category is wrong
a^2-[(a-b)^2+ (a-b)(a+b)] =
thank you!
Algebra.Com
Question 237617: Im sorry if the category is wrong
a^2-[(a-b)^2+ (a-b)(a+b)] =
thank you!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a^2-[(a-b)^2+ (a-b)(a+b)]
-------
= a^2 - [ (a^2-2ab+b^2) + (a^2-b^2)]
---
= a^2 - [2a^2 -2ab]
---
= 2ab - a^2
=====================
Cheers,
Stan H.
RELATED QUESTIONS
I'm sorry if I put this on wrong category :(
What is the integral [from b to a] 2/x^3?... (answered by Alan3354)
My question is about factoring. If factoring is not considered an operation on exponents, (answered by ankor@dixie-net.com)
A 500 lira coin is dropped from the leaning tower of pisa.It starts from rest and falls... (answered by richwmiller)
The question on my homework asks to "graph each function rule"
1. y=2-x
I'm pretty... (answered by rfer)
If a, b, c are odd integers, then the equation {{{ ax^2 + bx + c = 0 }}} has no integer... (answered by jim_thompson5910)
Sorry Wrong question
With the equation {{{ 8^2-9=0 }}} does a=8 b=0 and c=9?... (answered by vertciel)
Please help me with the following..Thank you!
Solve for x in... (answered by god2012)
If {{{ a + b = a/b + b/a }}} where a and b are positive integers, find the value of {{{... (answered by math_helper,Edwin McCravy)
If a, b, c are odd integers, then the equation {{{ ax^2+bx+c=0 }}} has no FRACTION... (answered by ikleyn,richard1234)