# [SOLVED] Simplification of Algebraic fractions

• May 9th 2009, 10:03 PM
waven
[SOLVED] Simplification of Algebraic fractions
Q. Simplify:

$\frac {(a + b)^2 - c^2} {3a + 3b - 3c}$

i've got up to

$\frac {a^2 + 2ab +b^2 - c^2} {3(a + b - c)}$

but i can't simplify it further unless $\frac {a^2 + 2ab +b^2 - c^2} {3(a + b - c)}$ is replaced by $\frac {a^2 + b^2 - c^2}{3(a + b - c)}$ then i could work my way to the answer, but im not sure if that should be the right method
because of the rule, $(a + b)^2 = a^2 + 2ab + b^2$

btw the answer is $\frac {a + b - c}3$
• May 10th 2009, 12:06 AM
TheEmptySet
Quote:

Originally Posted by waven
Q. Simplify:

$\frac {(a + b)^2 - c^2} {3a + 3b - 3c}$

i've got up to

$\frac {a^2 + 2ab +b^2 - c^2} {3(a + b - c)}$

but i can't simplify it further unless $\frac {a^2 + 2ab +b^2 - c^2} {3(a + b - c)}$ is replaced by $\frac {a^2 + b^2 - c^2}{3(a + b - c)}$ then i could work my way to the answer, but im not sure if that should be the right method
because of the rule, $(a + b)^2 = a^2 + 2ab + b^2$

btw the answer is $\frac {a + b - c}3$

Note the numerator is the difference of squares

$\frac {(a + b)^2 - c^2} {3a + 3b - 3c}=\frac{(a+b-c)(a+b+c)}{3(a+b-c)}=\frac{a+b+c}{3}$