Factorise:
@^
\begin{aligned}
1 - 2cd - c^2 - d^2
\end{aligned}
@^
Answer:
^@ (1 - c - d) (1 + c + d) ^@
- We know that @^ \begin{aligned} & (a^2 - b^2) = (a + b) (a - b) && \ldots \text{(i)} \\ & ( a + b )^2 = a^2 + 2ab + b^2 && \ldots \text{(ii)} \end{aligned} @^
- We have: @^ \begin{aligned} 1 - 2cd - c^2 - d^2 = \space& 1 - (c^2 + d^2 + 2cd )\\ = \space& 1^2 - (c + d)^2 && \text{[Using } eq \text{(ii)]} \\ = \space& (1 - (c + d) ) (1 + (c + d) ) && \text{[Using } eq \text{(i)]}\\ = \space& (1 - c - d) (1 + c + d) \end{aligned} @^
- Hence, ^@ 1 - 2cd - c^2 - d^2 = \space (1 - c - d) (1 + c + d). ^@
