Question

If (2p + 3q = 10) and (8p3 + 27q3 = 100), find the value of pq.
__________________________
Please don't spam
Unnecessary answer will be REPORTED​

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

Step-by-step explanation:

2x – 3y = 10 and xy 16

2. – 3y = 10

Cubing both the sides,

= 103

= (2x)3 – (34)3 – 3(2x)(3y)(2x – 3y) = 1000

[Since

(a – b)3 = a3 – 63 – 3ab(a – b)] = (2x)3 – (3y)3 – 18zy(2x – 3y) = 1000

[Since xy 16 and 2x 3y = 10]

> 8x3 – 27y3 – - 18 x 16 x 10 = 1000

» 8x3 27y – 2880 1000 = = 8x 27y3 : 1000 + 2880

→ 8x3 – 27y3 = 3880