question_answer 1)

                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) \[(a-b)\,(a+b)\,+(b-c)\,(b+c)\,+(c-a)\]\[(c+a)=0\]

Answer:

                (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.