Question

x+y+z =6 and xy+yz+zx=11 than value x³+y³+z³-3xyz​

Answer 1

2+2+2=6

x3+y3+z3 -3xyz

(2)cube (2)cube (2)cube

8+8+8

24 is answer now subtract with 3xyz(xyz value is 2)

24-18

4 is the answer

thank you

Answer 2

x+y+z= 6 ........ equation 1

xy+yz+zx =11 .....equation 2

the formula for x^3 +y^3 +z^3 -3xyz is : (x+y+z) (x^2+y^2+z^2-xy-yz-zx)

now putting values in our formula

=(x+y+z) (x^2 + y^2 + z^2 -(xy+yz+zx)) ......equation 3

Here first we will find the value of (x^2 + y^2 + z^2)

squaring both sides of equation 1...

(x+y+z)^2 = (6)^2

x^2 +y^2 +z^2 +2xy +2yz +2zx = 36

x^2 + y^2 + z^2 + 2(xy+yz+zx) =36

x^2 + y^2 +z^2 + 2*11 = 36

x^2 +y^2 + z^2 = 36-22

x^2 + y^2 +z^2 = 14

Now putting all values in equation 3,

= (6) (14-11)

= 6 × 3

= 18

The value of x^3 + y^3 + z^3 - 3xyz is 18..

hope you get help from this....

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