Question

solve the following equation and find x^2+ y^2
x^3 + 3xy^2 =14
y^3 + 3x^2y= 13​​

Answer 1

Required Answer:-

Given Information:

• x³ + 3xy² = 14
• y³ + 3x²y = 13

To find:

• The value of x² + y²

Solution:

Given,

➡ x³ + 3xy² = 14 ....(i)

➡ y³ + 3x²y = 13 ....(ii)

Adding both equations (i) and (ii), we get,

➡ x³ + y³ + 3x²y + 3xy² = 14 + 13

➡ (x + y)³ = 27

➡ x + y = 3 ......(iii)

Subtracting equation (ii) from (i), we get,

➡ x³ - y³ - 3x²y + 3xy² = 1

➡ (x - y)³ = 1

➡ x - y = 1 ........(iv)

Adding both the equations (iii) and (iv), we get,

➡ 2x = 4

➡ x = 2

From (iii), y = x - 1

➡ y = 1

So,

x = 2 and y = 1

Therefore,

x² + y²

= 2² + 1²

= 4 + 1

= 5

Hence,

➡ x² + y² = 5 (Answer)

Answer:

• x² + y² = 5

Identity Used:

➡ (x + y)³ = x³ + y³ + 3x²y + 3xy²

➡ (x - y)³ = x³ - y³ - 3x²y + 3xy²

Other Identities:

➡ (x + y)² = x² + 2xy + y²

➡ (x - y)² = x² - 2xy + y²

➡ (x + y)² = (x - y)² + 4ab

➡ (x - y)² = (x + y)² - 4ab