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(a-b+c-d)(a+b-c+d)

 Oct 13, 2016

Best Answer 

 #2
avatar+26387 
+5

(a-b+c-d)(a+b-c+d)

\(\begin{array}{|rcll|} \hline && (a-b+c-d)(a+b-c+d) \\ &=& [a-(b-c+d)][a+(b-c+d)] \qquad | \qquad (u-v)(u+v) = u^2-v^2 \\ &=& a^2-(b-c+d)^2 \\ &=& a^2-(b-c+d)(b-c+d) \\ &=& a^2-[(b-c)+d][(b-c)+d] \qquad | \qquad (u+v)(u+v) = u^2+2uv+v^2\\ &=& a^2-[ (b-c)^2+2(b-c)d+d^2 ] \\ &=& a^2-(b-c)^2-2(b-c)d-d^2 \\ &=& a^2-(b-c)^2-2bd+2cd-d^2 \qquad | \qquad (u-v)(u-v) = u^2-2uv+v^2\\ &=& a^2-(b^2-2bc+c^2)-2bd+2cd-d^2 \\ &=& a^2-b^2+2bc-c^2-2bd+2cd-d^2 \\ &=& a^2-b^2-c^2-d^2+2bc-2bd+2cd \\ \hline \end{array}\)

 

laugh

 Oct 13, 2016
 #1
avatar
+5

Expand the following:
(a-b+c-d) (a+b-c+d)

(a-b+c-d) (a+b-c+d) = a (a-b+c-d)+b (a-b+c-d)-c (a-b+c-d)+(a-b+c-d) d:
a (a-b+c-d)+b (a-b+c-d)-c (a-b+c-d)+d (a-b+c-d)

a (a-b+c-d) = a a+a (-b)+a c+a (-d):
a a-a b+a c-a d+b (a-b+c-d)-c (a-b+c-d)+d (a-b+c-d)

a a = a^2:
a^2-a b+a c-a d+b (a-b+c-d)-c (a-b+c-d)+d (a-b+c-d)

b (a-b+c-d) = b a+b (-b)+b c+b (-d):
a^2-a b+a c-a d+b a-b b+b c-b d-c (a-b+c-d)+d (a-b+c-d)

b (-b) = -b^2:
a^2-a b+a c-a d+b a+-b^2+b c-b d-c (a-b+c-d)+d (a-b+c-d)

-c (a-b+c-d) = -(c a)-c (-b)-c c-c (-d):
a^2-a b+a c-a d+b a-b^2+b c-b d+-(c a)-(-c b)-c c-(-c d)+d (a-b+c-d)

-(-1) = 1:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c c-(-c d)+d (a-b+c-d)

-c c = -c^2:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b+-c^2-(-c d)+d (a-b+c-d)

-(-1) = 1:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d (a-b+c-d)

d (a-b+c-d) = d a+d (-b)+d c+d (-d):
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d a-d b+d c-d d

d (-d) = -d^2:
a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d a-d b+d c+-d^2

Grouping like terms, a^2-a b+a c-a d+b a-b^2+b c-b d-c a+c b-c^2+c d+d a-d b+d c-d^2 = -d^2+(c d+c d)+(-(b d)-b d)-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b):
-d^2+(c d+c d)+(-(b d)-b d)-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

c d+c d = 2 (c d):
-d^2+2 c d+(-(b d)-b d)-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

-(b d)-b d = -2 (b d):
-d^2+2 c d+-2 b d-c^2+(b c+b c)-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

b c+b c = 2 (b c):
-d^2+2 c d-2 b d-c^2+2 b c-b^2+a^2+(a d-a d)+(a c-a c)+(a b-a b)

a d-a d = 0:
-d^2+2 c d-2 b d-c^2+2 b c-b^2+a^2+(a c-a c)+(a b-a b)

a c-a c = 0:
-d^2+2 c d-2 b d-c^2+2 b c-b^2+a^2+(a b-a b)

a b-a b = 0:
Answer: |-d^2 + 2cd - 2bd - c^2 + 2bc - b^2 + a^2

 Oct 13, 2016
 #2
avatar+26387 
+5
Best Answer

(a-b+c-d)(a+b-c+d)

\(\begin{array}{|rcll|} \hline && (a-b+c-d)(a+b-c+d) \\ &=& [a-(b-c+d)][a+(b-c+d)] \qquad | \qquad (u-v)(u+v) = u^2-v^2 \\ &=& a^2-(b-c+d)^2 \\ &=& a^2-(b-c+d)(b-c+d) \\ &=& a^2-[(b-c)+d][(b-c)+d] \qquad | \qquad (u+v)(u+v) = u^2+2uv+v^2\\ &=& a^2-[ (b-c)^2+2(b-c)d+d^2 ] \\ &=& a^2-(b-c)^2-2(b-c)d-d^2 \\ &=& a^2-(b-c)^2-2bd+2cd-d^2 \qquad | \qquad (u-v)(u-v) = u^2-2uv+v^2\\ &=& a^2-(b^2-2bc+c^2)-2bd+2cd-d^2 \\ &=& a^2-b^2+2bc-c^2-2bd+2cd-d^2 \\ &=& a^2-b^2-c^2-d^2+2bc-2bd+2cd \\ \hline \end{array}\)

 

laugh

heureka Oct 13, 2016

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