Question

Show that :
$(3xy-7yz{)}^2+84x{y}^{2}z=(3xy+7yz{)}^2$

Answer 1

$L.H.S$

$(3xy-7yz{)}^2+84x{y}^{2}z$

$=9x^2{y}^2+49y^2{z}^2-2\times 3\times 7x{y}^{2}z+84x{y}^{2}z$ $[\because (a-b{)}^2=a^2+b^2-2ab]$

$=9x^2{y}^2+49y^2{z}^2-42x{y}^{2}z+84x{y}^{2}z$

$=9x^2{y}^2+49y^2{z}^2+42x{y}^{2}z$

$=(3xy+7yz{)}^2$ $[\because (a+b{)}^2=a^2+b^2+2ab]$

$=R.H.S$

Hence proved.