Question

4 x + 3 y square - 4 x minus 3 y whole square minus 48 x y​

Answer 1

$\textbf{Given:}$

$(4x+3y)^2-(4x-3y)^2-48\,xy$

$\textbf{To simplify:}$

$(4x+3y)^2-(4x-3y)^2-48\,xy$

$\textbf{Solution:}$

$\text{Consider,}$

$(4x+3y)^2-(4x-3y)^2-48\,xy$

$\text{Using the identity,}$

$\boxed{\bf\,a^2-b^2=(a-b)(a+b)}$

$=((4x+3y)-(4x-3y))((4x+3y)+(4x-3y))-48\,xy$

$=(4x+3y-4x+3y)(4x+3y+4x-3y)-48\,xy$

$=(6y)(8x)-48\,xy$

$=48\,xy-48\,xy$

$=0$

$\textbf{Answer:}$

$\textbf{The value of \bf\,(4x+3y)^2-(4x-3y)^2-48\,xy is 0}$

Answer 2

olution:

\text{Consider,}Consider,

(4x+3y)^2-(4x-3y)^2-48\,xy(4x+3y)

2

−(4x−3y)

2

−48xy

\text{Using the identity,}Using the identity,

\boxed{\bf\,a^2-b^2=(a-b)(a+b)}

a

2

−b

2

=(a−b)(a+b)

=((4x+3y)-(4x-3y))((4x+3y)+(4x-3y))-48\,xy=((4x+3y)−(4x−3y))((4x+3y)+(4x−3y))−48xy

=(4x+3y-4x+3y)(4x+3y+4x-3y)-48\,xy=(4x+3y−4x+3y)(4x+3y+4x−3y)−48xy

=(6y)(8x)-48\,xy=(6y)(8x)−48xy

=48\,xy-48\,xy=48xy−48xy

=0=0