Answer :

_{Formula used:}

_{(a + b)(a – b) = a}^{2} – b^{2}

_{i. (xy + pq)(xy – pq)}

Taking a = xy and b = pq, then from above identity,

= (xy)^{2} – (pq)^{2}

= x^{2}y^{2} – p^{2}q^{2}

ii. 49 × 51

= (50 – 1)(50 + 1)

Taking a = 50 and b = 1, then from above identity,

= 50^{2} - 1^{2}

= 2500 – 1

= 2499

iii. (2x – y + 3z)(2x + y + 3z)

Taking a = 2x and b = y + 3z, the from the above identity we have

= (2x)^{2} – (y + 3z)^{2}

= 4x^{2} – (y^{2} + 2(y)(3z) + (3z)^{2}) [∵ (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4x^{2} – (y^{2} + 6yz + 9z^{2})

= 4x^{2} – y^{2} – 6yz – 9z^{2}

iv. 1511 × 1489

= (1500 + 11)(1500 – 11)

Taking a = 1500 and b = 11, then from above identity,

= (1500)^{2} - 11^{2}

= 2250000 – 121

= 2249879

v. (a – 2)(a + 2)(a^{2} + 4)

= (a^{2} – 2^{2})(a^{2} + 4) [Taking a = a, and b = 2 in above identity]

= (a^{2} – 4)(a^{2} + 4)

= (a^{2})^{2} - 4^{2} [Taking a = a^{2} and b = 4 in above identity]

= a^{4} – 16

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