Factorise:
1−2cd−c2−d2
Answer:
(1−c−d)(1+c+d)
- We know that (a2−b2)=(a+b)(a−b)…(i)(a+b)2=a2+2ab+b2…(ii)
- We have: 1−2cd−c2−d2= 1−(c2+d2+2cd)= 12−(c+d)2[Using eq(ii)]= (1−(c+d))(1+(c+d))[Using eq(i)]= (1−c−d)(1+c+d)
- Hence, 1−2cd−c2−d2= (1−c−d)(1+c+d).