Question

Show that (7a+4xy)2-(7a-4xy)2=56axy

Answer 1

Hey friend , Harish here.
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Here is your answer
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We know that , 

$(a + b)^{2} = a^{2} +b^{2} + 2 ab$

$(a-b)^{2} = a^{2}+b^{2}-2ab$

Now, use the above given identities .

Then, 

$(7a+4xy) ^{2} = (7a)^{2} + (4xy)^{2} + 2(7a)(4xy)$

                      $= 49a^{2} + 16 x^{2}y^{2} + 56 axy$ 


$(7a-4xy)^{2}= (7a)^{2}+(4xy)^{2} - 2(7a)(4xy)$

                   $= 49a^{2}+16 x^{2} y^{2} - 56a xy$

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Now, We Should simply & prove LHS = RHS.

$( 7a +4xy)^{2} - ( 7a -4xy)^{2} = 112axy.$

=> $(49a^{2} + 16 x^{2}y^{2} + 56 axy )-(49a^{2} + 16 x^{2}y^{2} - 56 axy) = 112 axy$

Now remove the brackets.

Then,

=> $(49a^{2} + 16 x^{2}y^{2} + 56 axy - 49a^{2} - 16 x^{2}y^{2} +56 axy) = 112 axy$

=>$56axy + 56 axy = 112axy$

=>   $112axy = 112 axy$

            $LHS=RHS$

Hence proved.
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Hope my answer is helpful to u.