Question

If x+y+z=12 and x square +y square + z square=70 then find the value of x cube+ycube+z cube -3xyz.

Answer quick...pleaseee

Answer 1

Answer:

x³+y³+z³-3xyz=803 is your answer mate

Step-by-step explanation:

Given that:

x+y+z=12

x²+y²+z²=70

x³+y³+z³-3xyz=?

solution:

we know that:

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²)-(xy+yz+zx)---(A)

we also know that:

(x+y+z)²=x²+y²+z²+2(xy+yz+zx)---(B)

putting all the values in eq (B) first

(12)²=70+2(xy+yz+zx)

(144-70)/2=xy+yz+zx

74/2= xy+yz+zx

xy+yz+zx=37

now putting All these values in eqn (A)

x³+y³+z³-3xyz=(12)(70)-(37)

x³+y³+z³-3xyz=840-37

x³+y³+z³-3xyz=803

Answer 2

Answer:

Step-by-step explanation: