Question

If (2p + 3q = 10) and (8p3 + 27q3 = 100), find the value of pq.
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Answer 1

Given that

$2p+3q=10$ ...[Equation 1]

$8p^3+27q^3=100$ ...[Equation 2]

Factorization of Equation 2

$\rightarrow (2p+3q)(4p^2-6pq+9p^2)=100$

$\rightarrow 10(4p^2-6pq+9q^2)=100$

$\rightarrow 4p^2-6pq+9q^2=10$ ...[Equation 3]

Since this question requires identities, for convenience, let the sum and product of $2p,3q$ be $m,n$.

Then Equation 1, 3 are

$\rightarrow \begin{cases} & m=10 \\ & m^2-3n=10 \end{cases}$

$\rightarrow m=10,n=30$

Hence $6pq=30$, and hence $pq=5$.

Answer 2

2p+3q=18⟹p=218−3q4p2+4pq−3q2−36=0 

⟹4(218−3q)2+4(218−3q)q−3q2−36=0⟹324−36=(108−36)q⟹72q=288⟹q=4⟹p=218−12=3⟹2p+q=2(3)+4=10

Hope's it helps you