• # question_answer 20)                 Find the following squares by using the identities.                 (i) ${{(b-7)}^{2}}$                          (ii) ${{(xy+3z)}^{2}}$                    (iii) ${{(6{{x}^{2}}-5y)}^{2}}$                     (iv) ${{\left( \frac{2}{3}\,m+\frac{3}{2}n \right)}^{3}}$                 (v) ${{(0.4\,p-0.5\,q)}^{2}}$      (vi) ${{(2xy\,+5y)}^{2}}$.

(i) ${{(b-7)}^{2}}$                          $=(b-7)\,(b-7)$ $=b(b-7)\,-7(b-7)$ $={{b}^{2}}-7b-7b+49$ $={{b}^{2}}-14b+49$ (ii) ${{(xy+3z)}^{2}}$$=(xy+3z)\,(xy+3z)$ $=xy\,(xy+3z)+3z(xy+3z)$ $={{x}^{2}}{{y}^{2}}\,+3xyz\,+3xyz\,+9{{z}^{2}}$ $={{x}^{2}}{{y}^{2}}\,+6xyz+9{{z}^{2}}$ (iii) ${{(6{{x}^{2}}-5y)}^{2}}\,=(6{{x}^{2}}-5y)\,(6{{x}^{2}}-5y)$ $=6{{x}^{2}}\,(6{{x}^{2}}-5y)\,-5y\,(6{{x}^{2}}-5y)$ $=36{{x}^{4}}\,-30{{x}^{2}}y\,-30{{x}^{2}}y\,+25{{y}^{2}}$ $=36{{x}^{4}}-60{{x}^{2}}y\,+25{{y}^{2}}$ (iv) ${{\left( \frac{2}{3}\,m+\frac{3}{2}n \right)}^{3}}$ $=\left( \frac{2}{3}m+\frac{3}{2}n \right)\,\left( \frac{2}{3}m+\frac{3}{2}n \right)$ $=\frac{2}{3}\,m\left( \frac{2}{3}m+\frac{3}{2}n \right)+\frac{3}{2}n\,\left( \frac{2}{3}m+\frac{3}{2}n \right)$ $=\frac{4}{9}{{m}^{4}}+mn+mn+\frac{9}{4}\,{{n}^{2}}$ $=\frac{4}{9}\,{{m}^{2}}+2\,mn\,+\frac{9}{4}{{n}^{2}}$                 (v) ${{(0.4\,p-0.5\,q)}^{2}}$ $=(0.4\,-0.5q)\,-(0.4p-0.5q)$ $=0.4p(0.4p-0.5q)-0.5q(0.4p-0.5q)$ $=0.16{{p}^{2}}\,-0.2pq-0.2pq\,+0.25{{q}^{2}}$ $=0.16{{p}^{2}}-0.4pq+0.25\,{{q}^{2}}$                 (vi) ${{(2xy\,+5y)}^{2}}$ $=(2xy\,+5y)\,\,(2xy+5y)$ $=2xy\,(2xy+5y)\,+5y\,(2xy+5y)$ $=4{{x}^{2}}{{y}^{2}}\,+10x{{y}^{2}}\,+10x{{y}^{2}}+25{{y}^{2}}$ $=4{{x}^{2}}{{y}^{2}}\,+20x{{y}^{2}}+25{{y}^{2}}$.