How do you solve:
(a+b)^2 - (b+c)^2 - (a+c)^2
2006-12-21 09:14:38 · 5 answers · asked by D M 1 in Science & Mathematics ➔ Mathematics
What does this equation equal?
2006-12-21 09:17:09 · update #1
Simplify the Question
2006-12-21 09:17:34 · update #2
It can be simplified. Use foil.
(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2
(b+c)^2 = (b+c)(b+c) = b^2 + 2bc + c^2
(a+c)^2 = (a+c)(a+c) = a^2 + 2ac + c^2
So, we have
(a+b)^2 - (b+c)^2 - (a+c)^2
= a^2 + 2ab + b^2 - (b^2 + 2bc + c^2) - (a^2 + 2ac + c^2)
= a^2 + 2ab + b^2 - b^2 - 2bc - c^2 - a^2 - 2ac - c^2
= 2ab - 2bc - 2ac - 2c^2
2006-12-21 09:16:33 · answer #1 · answered by MsMath 7 · 1⤊ 0⤋
(A+B)^2 = (A+B)(A+B) = A^2 + 2. A. B + B^2
Applying this relation to all 3 binomial squares,
(a+b)^2 - (b+c)^2 - (a+c)^2
= (a^2 + 2ab + b^2) - (b^2 + 2bc + c^2) - (a^2 + 2ac + c^2)
= a^2 + 2ab + b^2 - b^2 - 2bc - c^2 - a^2 - 2ac - c^2
= 2ab - 2bc - 2ac - 2c^2
2006-12-21 18:29:00 · answer #2 · answered by Sheen 4 · 1⤊ 0⤋
(a+b)^2 - (b+c)^2 - (a+c)^2=
a^2 + 2ab + b^2 - b^2 - 2bc - c^2 - a^2 -2ac - c^2=
2ab - 2bc - 2ac - 2c^2=
2(ab - bc - ac - c^2)
or 2(ab - c(b - a - c))
this is the best i could make out of this equation, hope that it helps
2006-12-21 17:22:05 · answer #3 · answered by b0b 7h3 l337 2 · 1⤊ 0⤋
You might mean simplify which would then be
(a^2+2ab+b^2)-(b^2+2bc+c^2)-(a^2+2ac+c^2)
a^2+2ab+b^2-b^2-2bc-c^2-a^2-2ac-c^2
2ab-2bc-2ac-2c^2
........?
2006-12-21 17:16:59 · answer #4 · answered by a_math_guy 5 · 1⤊ 0⤋
(a+b)² - (b+c)² - (a+c)²
(a² + 2ab + b²)-(b² + 2bc + c²)-(a² + 2ac+c²) =
(2ab)-(2bc)-(2ac) =
2(ab - bc - ac)
::
2006-12-21 17:28:44 · answer #5 · answered by aeiou 7 · 0⤊ 1⤋