Question

if x+y+z=8 and xy+yz+zx=20 then find the value of x cube+ y cube + z cube- 3xyz​

Answer 1

Answer:

Given, x+y+z=8,xy+yz+zx=20

Now, x³+y³+z³ −3xyz=(x+y+z)(x²+y²+z²−xy−yz−zx)

Again, (x+y+z)²=x²+y²+z²+2xy+2yz+2zx

x²+y²+z² =(x+y+z)²-2(xy+yz+zx)

x²+y²+z²=(8)²-2(20)=24

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

x³+y³+z³−3xyz=(8)[24−(20)]

x³+y³+z³-3xyz=(8)[24−(20)]=8(4)=32