Question

if 2x+3y=7 and xy=2,find 8xcube +27ycube​

Answer 1

Required Answer:-

Given:

$\begin{cases} \sf 2x + 3y = 7 \\ \sf xy = 2 \end{cases}$

To Find:

• 8x³ + 27y³ = ?

Solution:

Given that,

$\sf \implies 2x + 3y = 7$

Cubing both sides, we get,

$\sf \implies (2x + 3y)^{3} = {7}^{3}$

Using identity (a + b)³ = a³ + b³ + 3ab(a + b), we get,

$\sf \implies 8 {x}^{3} + 27 {y}^{3} + 3 \times 2x \times 3y(2x + 3y)= {7}^{3}$

$\sf \implies 8 {x}^{3} + 27 {y}^{3} + 18xy(2x + 3y)= 343$

Substituting the values of xy and 2x + 3y, we get,

$\sf \implies 8 {x}^{3} + 27 {y}^{3} + 18 \times 2 \times 7= 343$

$\sf \implies 8 {x}^{3} + 27 {y}^{3} + 36 \times 7= 343$

$\sf \implies 8 {x}^{3} + 27 {y}^{3} =7 \times 49 - 7 \times 36$

$\sf \implies 8 {x}^{3} + 27 {y}^{3} =7 \times( 49 -36)$

$\sf \implies 8 {x}^{3} + 27 {y}^{3} =7 \times13$

$\sf \implies 8 {x}^{3} + 27 {y}^{3} = 91$

→ Hence, the value of 8x³ + 27y³ is 91.

Answer:

• 8x³ + 27y³ = 91

More Identities to know:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)² = (a - b)² + 4ab
• (a - b)² = (a + b)² - 4ab
• (a + b)² + (a - b)² = 2(a² + b²)
• (a + b)³ = a³ + b³ + 3ab(a + b)
• (a - b)³ = a³ - b³ - 3ab(a - b)