• # question_answer 22)                 Show that                 (i) ${{(3x+7)}^{2}}-84x={{(3x-7)}^{2}}$                 (ii) ${{(9p-5q)}^{2}}+180pq={{(9p+5q)}^{2}}$                 (iii) ${{\left( \frac{4}{3}m-\frac{3}{4}n \right)}^{2}}+2mn=\frac{16}{9}\,{{m}^{2}}+\frac{9}{16}{{n}^{2}}$                 (iv) ${{(4pq+3q)}^{2}}-{{(4pq-3q)}^{2}}=48p{{q}^{2}}$                 (v) $(a-b)\,(a+b)\,+(b-c)\,(b+c)\,+(c-a)$$(c+a)=0$

(i) ${{(3x+7)}^{2}}-84x={{(3x-7)}^{2}}$                 L.H.S. $={{(3x+7)}^{2}}-84x$ $=\{{{(3x)}^{2}}+2(3x)\,(7)+{{(7)}^{2}}\}\,-84x$ $=(9{{x}^{2}}+42x+49)\,-84x$ $=9{{x}^{2}}+(42x-84x)+49$                     |Combining the like terms $=9{{x}^{2}}-42x+49$                  ?(1) R.H.S. $={{(3x-7)}^{2}}$ $={{(3x)}^{2}}-2(3x)\,(7)\,+{{(7)}^{2}}$ $=9{{x}^{2}}-42x+49$                  ?(2) From equations (1) and (2), ${{(3x+7)}^{2}}-84x={{(3x-7)}^{2}}$                 (ii) ${{(9p-5q)}^{2}}+180pq={{(9p+5q)}^{2}}$ L.H.S.$={{(9p-5q)}^{2}}+180pq$ $=\{{{(9p)}^{2}}-2(9p)\,(5q)+{{(5q)}^{2}}\}\,+180pq$ $=(81{{p}^{2}}-90pq+25{{q}^{2}})\,+180pq$ $=81{{p}^{2}}+(180pq-90pq)+25{{q}^{2}}$                         |Combing the like terms $=81{{p}^{2}}+90pq+25{{q}^{2}}$                          ?(1) R.H.S.    $={{(9p+5q)}^{2}}$ $={{(9p)}^{2}}+2(9p)(5q)\,+{{(5q)}^{2}}$ $=81{{p}^{2}}+9pq+25{{q}^{2}}$                            ?(2)                 From equations. (1) and (2) ${{(9p-5q)}^{2}}+180pq={{(9p+5q)}^{2}}$                 (iii) ${{\left( \frac{4}{3}m-\frac{3}{4}n \right)}^{2}}+2mn=\frac{16}{9}\,{{m}^{2}}+\frac{9}{16}{{n}^{2}}$                 L.H.S. $={{\left( \frac{4}{3}m-\frac{3}{4}n \right)}^{2}}+2mn$                 $={{\left( \frac{4}{3}m \right)}^{2}}\,-2\left( \frac{4}{3}m \right)\left( \frac{3}{4}m \right)+{{\left( \frac{3}{4}m \right)}^{2}}+2mn$ $=\frac{16}{9}{{m}^{2}}-2mn+\frac{9}{16}\,{{n}^{2}}+2mn$ $=\frac{16}{9}\,{{m}^{2}}+(2mn-2mn)\,+\frac{9}{16}\,{{n}^{2}}$                            |Combining the like terms $=\frac{16}{9}{{m}^{2}}+\frac{9}{16}\,{{n}^{2}}$ = R.H.S. (iv) $(4pq)+3q{{)}^{2}}-{{(4pq-3q)}^{2}}=48p{{q}^{2}}$ L.H.S.     $={{(4pq+3q)}^{2}}-{{(4pq-3q)}^{2}}$ $=\{{{(4pq)}^{2}}+2(4pq)\,(3q)+{{(3q)}^{2}}\}$$-\{{{(4pq)}^{2}}-2(4pq)\,(3q)\,+{{(3q)}^{2}}\}$ $=(16{{p}^{2}}{{q}^{2}}+24p{{q}^{2}}+9{{q}^{2}})$$-(16{{p}^{2}}{{q}^{2}}-24p{{q}^{2}}+9{{q}^{2}})$ $=16{{p}^{2}}{{q}^{2}}+24p{{q}^{2}}+9{{q}^{2}}-16{{p}^{2}}{{q}^{2}}$$+24p{{q}^{2}}-9{{q}^{2}}$ $=(16{{p}^{2}}{{q}^{2}}-16{{p}^{2}}{{q}^{2}})+(24p{{q}^{2}}+24p{{q}^{2}})$$+(9{{q}^{2}}-9{{q}^{2}})$                |Combining the like terms                 $=48p{{q}^{2}}$                 = R.H.S.                 (v) $+(c-a)\,(c+a)=0$                 L.H.S. $=(a-b)\,(a+b)\,+(b-c)\,(b+c)$$+(c-a)\,(c+a)$                 $={{a}^{2}}-{{b}^{2}}+{{b}^{2}}-{{c}^{2}}+{{c}^{2}}-{{a}^{2}}$                    |Using identity III $=({{a}^{2}}-{{a}^{2}})+({{b}^{2}}-{{b}^{2}})+({{c}^{2}}-{{c}^{2}})$          |Combining the like terms = 0 = R.H.S.