Question
Verify the following: $(7p - 13q)^2+ 364pq = (7p + 13q)^2$
✓
Answer
$(7p - 13q)^2+ 364pq = (7p + 13q)^2$
Taking $LHS$
$ =(7 p-13 q)^2+364 p q $
$ =(7 p)^2+(13 q)^2-2 \times 7 p \times 3 q+364 p q $
$ =(7 p)^2+(13 q)^2-182 p q+364 p q $
$ =(7 p)^2+(13 q)^2+182 p q $
$ =(7 p)^2+(13 q)^2+2 \times 7 p \times 13 q $
$=(7 p+13 q)^2 $
$= RHS$
Hence verified
Taking $LHS$
$ =(7 p-13 q)^2+364 p q $
$ =(7 p)^2+(13 q)^2-2 \times 7 p \times 3 q+364 p q $
$ =(7 p)^2+(13 q)^2-182 p q+364 p q $
$ =(7 p)^2+(13 q)^2+182 p q $
$ =(7 p)^2+(13 q)^2+2 \times 7 p \times 13 q $
$=(7 p+13 q)^2 $
$= RHS$
Hence verified
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