Question

12. If x + y + z = 8 and xy + yz + zx = 20, find the value of x3 + y3 + z3 – 3xyz.

Answer 1

We know that,
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
X^2+y^2+z^2=(8)^2-2(20)
=64-40
=24
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=(8)(24-20)
=8*4
=32

Answer 2

Answer:

The value is 32

Step-by-step explanation:

Given the value of

$x + y + z = 8$

$xy + yz + zx = 20$

$\text{we have to find the value of }x^3+y^3+z^3-3xyz$

By the identity

$a^3 + b^3 + c^3 - 3abc = (a + b + c )(a^2 + b^2 + c^2 -ab - ac -bc)$

$a^3 + b^3 + c^3 - 3abc =(a+b+c)\{(a^2+b^2+c^2+2ab+2bc+2ac)- 3(ab+bc+ac)\}$

$a^3 + b^3 + c^3 - 3abc = (a+b+c)\{(a+b+c)^2 - 3(ab+bc+ac)\}$  

Put a=x, b=y, c=z

$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\{(x+y+z)^2 - 3(xy+yz+zx)\}$ $x^3 + y^3 + z^3 - 3xyz=8(64-3(20))=8(64-60)=8\times 4=32$

Hence, the value is 32