Question

If x-y = 6 and xy = 10 , find the value of x³ + y³ .

Answer 1

Question :

If x-y = 6 and xy = 10 , find the value of x³+y³ .

Solution :

Given :
x-y =6

Taking cube on both sides , we get

(x-y)³ = 6³

x³-y³-3x²y+3xy² = 216

x³-y³-3xy (x-y) = 216

x³-y³ - 3xy (6) = 216 (x-y = 6)

x³-y³ -3 × 10 (6) = 216 (xy = 10)

x³-y³ -180 = 216

x³-y³ = 216+180 = 396

The required value of x³-y³ = 396

Answer 2

Answer:

x³ - y³ = 396, if x-y = 6 and xy = 10.

Step-by-step explanation:

Here, it is given that x-y = 6. Cubing both sides in this equation, we have, (x-y)³ = 6³. Opening the brackets and expanding, we have, x³ - y³ - 3xy(x-y) = 216. Also, they value for xy is given as 10. So, x³ - y³ - 3(10)(6) = 216. Multiplying 3*10*6, we have x³ - y³ - 180 = 216. From here, we can have the value for (x³-y³), ie, 216 + 180 = 396.

Thus, x³ - y³ = 396, if x-y = 6 and xy = 10.