If 3p-4q=5 and pq=3 . Find 27p^3-64q^3
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it is thw identity of a³-b³..
a³-b³= (a-b)(a²+b²-ab)
=(3p-4q){(3p)²+(4q)²-3p×4q}
=5(9p²+16q²-12pq)
=5{(3p-4q)²+24pq-12pq}
= 5(25+12pq)
=5(25+12×3)
=5×61=305
a³-b³= (a-b)(a²+b²-ab)
=(3p-4q){(3p)²+(4q)²-3p×4q}
=5(9p²+16q²-12pq)
=5{(3p-4q)²+24pq-12pq}
= 5(25+12pq)
=5(25+12×3)
=5×61=305
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