Question

If x+y+z=19, xyz= 216 and xy+ yz+ zx = 144, then the value of √x³+y³+z³+xyz is

Answer 1

Given :

• x + y + z = 19.
• xy + yz + zx = 144.
• xyz = 216.

To Find:

• The value of √x³+y³+z³ +xyz.

Answer:

Now here given that ,

=> (x+y+z) = 19.

=> (x+y+z)² = 19².

=> x²+y²+z²+2(xy+yz+zx) = 361.

=> x²+y²+z² +2(144) = 361.

=> x²+y²+z² = 361 - 288.

=> x²+y²+z² = 73.

______________________________________

Also we know a formula ,

→ a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca).

Till now we know that

1. x²+y²+z² = 73.
2. x + y + z = 19.
3. xy + yz + zx = 144.
4. xyz = 216.

Substituting the respective values,

=>x³+y³+z³-3×216=(19)×[73-(144)]

=> x³+y³+z³-648= 19× (-71).

=> x³+y³+z³ = -1349+648

=> x³+y³+z³=701.

_____________________________________

Hence √x³+y³+z³+xyz

= √ 701 + 216

= √ 917

= 30.28 ≈ 30