• question_answer 18)                 Use a suitable identity to get each of the following products: (i) $(x+3)(x-3)$ (ii) $(2y+5)(2y+5)$ (iii) $(2a-7)(2a-7)$  (iv) $\left( 3a-\frac{1}{2} \right)\,\left( 3a-\frac{1}{2} \right)$ (v) $(1.1\,m\,-\,0.4)\,(1.1\,m\,+\,0.4)$ (vi) $({{a}^{2}}+{{b}^{2}})\,(-{{a}^{2}}+{{b}^{2}})$ (vii) $(6x-7)\,(6x+7)$ (viii) $(-a+c)\,(-a+c)$ (ix) $\left( \frac{x}{2}+\frac{3y}{4} \right)\,\left( \frac{x}{2}\,+\frac{3y}{4} \right)$ (x) $(7a-9b)\,(7a-9b)$.

(i) $(x+3)\,(x+3)$                 $(x+3)\,(x+3)={{(x+3)}^{2}}$                 ${{(x)}^{2}}+2(x)\,(3)+{{(3)}^{2}}$                         |Using Identity I $={{x}^{2}}+6x+9$ (ii) $(2y+5)\,(2y+5)$ $(2y+5)\,(2y+5)$$={{(2y+5)}^{2}}$ ${{(=2y)}^{2}}+2(2y)(5)+{{(5)}^{2}}$     |Using Identity I $=4{{y}^{2}}+20y+25$  (iii) $(2a-7)\,(2a-7)$ $(2a-7)\,(2a-7)$ $={{(2a-7)}^{2}}$ $={{(2a)}^{2}}-2(2a)\,(7)+{{(7)}^{2}}$   |Using Identity II $=4{{a}^{2}}-28a+49$ (iv) $\left( 3a-\frac{1}{2} \right)\,\left( 3a-\frac{1}{2} \right)$ $\left( 3a-\frac{1}{2} \right)\,\left( 3a-\frac{1}{2} \right)$$={{\left( 3a-\frac{1}{2} \right)}^{2}}$ $={{(3a)}^{2}}-2(3a)\,\left( \frac{1}{2} \right)+{{\left( \frac{1}{2} \right)}^{2}}$ |Using Identity II $=9{{a}^{2}}-3a+\frac{1}{4}$ (v) $(1.1m-0.4)\,(1.1\,m+0.4)$ $(1.1m-0.4)\,(1.1\,m+0.4)$ $={{(1.1\,m)}^{2}}-0.16$                            |Using Identity III $=1.21\,{{m}^{2}}-0.16$ (vi) $({{a}^{2}}+{{b}^{2}})\,(-{{a}^{2}}+{{b}^{2}})$ $({{a}^{2}}+{{b}^{2}})\,(-{{a}^{2}}+{{b}^{2}})$$=({{b}^{2}}+{{a}^{2}})\,({{b}^{2}}-{{a}^{2}})$ $={{({{b}^{2}})}^{2}}\,-{{({{a}^{2}})}^{2}}$                                         |Using Identity III $={{b}^{4}}-{{a}^{4}}$ (vii) $(6x-7)\,(6x+7)$ $(6x-7)\,(6x+7)$$={{(6x)}^{2}}-{{(7)}^{2}}$     |Using  Identity III $=36{{x}^{2}}-49$ (viii) $(-a+c)\,(-a+c)$ $(-a+c)\,(-a+c)$$={{(-a+c)}^{2}}$ $={{(c-a)}^{2}}$ $={{c}^{2}}-2ca+{{a}^{2}}$                                         |Using Identity II (ix) $\left( \frac{x}{2}+\frac{3y}{4} \right)\,\left( \frac{x}{2}+\frac{3y}{4} \right)$ $\left( \frac{x}{2}+\frac{3y}{4} \right)\,\left( \frac{x}{2}+\frac{3y}{4} \right)$ $={{\left( \frac{x}{2} \right)}^{2}}+2\left( \frac{x}{2} \right)\,\left( \frac{3y}{4} \right)\,+{{\left( \frac{3y}{4} \right)}^{2}}$    |Using Identity I $=\frac{{{x}^{2}}}{4}+\frac{3xy}{4}\,+\frac{9{{y}^{2}}}{16}$ (x) $(7a-9b)\,(7a-9b)$ $(7a-9b)\,(7a-9b)$$={{(7a-9b)}^{2}}$ $={{(7a)}^{2}}-2(7a)(9b)\,+{{(9b)}^{2}}$             |Using Identity II $=49{{a}^{2}}-126ab+81{{b}^{2}}$.