Question 54

If $$x + y + z = 19, x^2 + y^2 + z^2 = 133$$, then the value of $$x^3 + y^3 + z^3 - 3xyz$$ is:

Solution

$$\left(x+y+z\right)=19$$

Squaring both side,

$$\left(x+y+z^{ }\right)^2=19^2$$

$$x^2+y^2+z^2+2\left(xy+yz+xz\right)=361$$

$$133+2\left(xy+yz+xz\right)=361$$

$$\left(xy+yz+xz\right)=114$$

$$x^3 + y^3 + z^3 - 3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz$$

$$x^3 + y^3 + z^3 - 3xyz=19 (133-114) = 361.

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