• # question_answer (a) Find the value of the expression $\left( 81{{x}^{2}}+16{{y}^{2}}-72xy \right),$ when $x=\text{ }\frac{2}{3}$and $y=\text{ }\frac{3}{4}$ (b) If a = 2 and b = 5, then verify ${{(a+b)}^{2}}=\text{ }{{a}^{2}}+{{b}^{2\text{ }}}+2ab.$

 (a)$81\text{ }{{x}^{2}}+16{{y}^{2}}-72xy={{\left( 9x \right)}^{2}}+{{\left( 4y \right)}^{2}}-2\times 9x\times 4y$ $={{(9x-4y)}^{2}}$ $[\because \,{{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}]$ Now, putting $x=\frac{2}{3}$and $y=\frac{3}{4},$then $={{\left( 9\times \frac{2}{3}-4\times \frac{3}{4} \right)}^{2}}$ $={{(6-3)}^{2}}={{3}^{2}}=9$ (b) Putting a = 2 and b = 5, then L.H.S $={{\left( a+b \right)}^{2}}$ $={{\left( 2+5 \right)}^{2}}={{7}^{2}}=49~$ and   R.H.S =$={{a}^{2}}+{{b}^{2}}+2ab$ $={{2}^{2}}+{{5}^{2}}+2\times 2\times 5$ = 4 + 25 + 20 = 49 Hence, L.H.S = R.H.S =49