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(4pq +3q)2²– (4pq – 3q)² = 48pq²
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(4pq + 3q)² – (4pq – 3q)² = 48pq²
As we know,
(a + b)² = a² + b² + 2ab
and
(a – b)² = a² + b² – 2ab
So, we have to apply the above identities in the given equation.
(4pq + 3q)² – (4pq – 3q)² = L.H.S.
= (16p²q² + 9q² + 24pq²) – (16p²q² + 9q² – 24pq²)
= 16p²q² + 9q² + 24pq² – 16p²q² – 9q² + 24pq²
= 24pq² + 24pq²
= 48pq²
= R.H.S.