Find the following squares by using the identities:
$( i)$. $( b-7)^{2}$
$( ii)$. $( xy+3z)^{2}$
$( iii)$. $( 6x^{2}-5 y)^{2}$
$( iv)$. $(\frac{2}{3}m+\frac{3}{2}n)^{2}$
$( v)$. $( 0.4p-0.5q)^{2}$
$( vi)$. $( 2xy+5y)^{2}$
Solution:
$( i)$. $( b-7)^2$
$=b^2+7^2-2( b)( 7)$
$=b^2+49-14b$ $\because [(a-b)^2=a^2+b^2-2ab]$
$( ii)$. $( xy+3z)^2$
$=( xy)^2+( 3z)^2+2( xy)( 3z)$
$=x^2y^2+9z^2+6xyz$ $\because [(a+b)^2=a^2+b^2+2ab]$
$( iii)$. $( 6x-5y)^2$
$=( 6x)^2+( 5y)^2-2( 6x)( 5y)$
$=36x^2+25y^2-60xy$ $\because [( a-b)^2=a^2+b^2-2ab]$
$( iv)$. $( \frac{2}{3}m+\frac{3}{2}n)^2$
$=( \frac{2}{3}m)^2+( \frac{3}{2}n)^2+2( \frac{2}{3}m)( \frac{3}{2}n)$
$=\frac{4}{9}m^2+\frac{9}{4}n^2+2mn$ $\because [(a+b)^2=a^2+b^2+2ab]$
$( v)$. $(0.4p-0.5q)^2$
$=( 0.4p)^2+( 0.5q)^2-2( 0.4p)( 0.5q)$
$=0.16p^2+0.25q^2-4pq$ $\because [( a-b)^2=a^2+b^2-2ab]$
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