Question

The value of${(\mathrm{a}+\mathrm{b})}^2–{(\mathrm{a}–\mathrm{b})}^2$ is

• A. $4\mathrm{ab}$
• B. $–4\mathrm{ab}$
• C. $2{\mathrm{a}}^2+2{\mathrm{b}}^2$
• D. $2{\mathrm{a}}^2–2{\mathrm{b}}^2$

Answer 1

The correct option is A

$4\mathrm{ab}$

Find the value of${(\mathrm{a}+\mathrm{b})}^2–{(\mathrm{a}–\mathrm{b})}^2$

$\begin{array}{rcl}{(\mathrm{a}+\mathrm{b})}^2& =& ({\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab})\mathrm{and}{(\mathrm{a}–\mathrm{b})}^2=({\mathrm{a}}^2+{\mathrm{b}}^2–2\mathrm{ab})\\ \mathrm{Then}{(\mathrm{a}+\mathrm{b})}^2–{(\mathrm{a}–\mathrm{b})}^2& =& ({\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab})–({\mathrm{a}}^2+{\mathrm{b}}^2–2\mathrm{ab})\\ & =& {\mathrm{a}}^2–{\mathrm{a}}^2+{\mathrm{b}}^2–{\mathrm{b}}^2+2\mathrm{ab}+2\mathrm{ab}\\ & =& 4\mathrm{ab}\\ \therefore {(\mathrm{a}+\mathrm{b})}^2–{(\mathrm{a}–\mathrm{b})}^2& =& 4\mathrm{ab}\end{array}$

$\therefore$The correct option is A.