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

Given:

• 2p+3q = 10
• 8p^3+27q^3 = 100

To find:

• The value of Pa

Solution:

⟹ 2p+3q = 10

⟹ (2p+3q)^3 = (10)^3

• We know that (a+b)^3 = a^3+b^3+3ab(a+b)

On using the identity,

⟹ 8p^3+27q^3+3×2p×3q(2p+3q) = 100

On substituting the values,

⟹ 100+18pq(10) = 100

⟹ 180pq= 0

⟹ pq= 0

Hence, the value of pq is 0.