Question

If x, y, and z are positive integers such that (x + y + z)2 = 16 and xy + yz + zx = 5, then what is the value of the expression x3 + y3 + z3 − 3xyz?

Answer 1

Answer:

I dont no because i read in sixth class

Answer 2

Answer:

4

Step-by-step explanation:

$(x+y+z)^2=16$

If we square root whole thing

$\sqrt{(x+y+z)^{2} } = \sqrt{16}$

which means

= x+y+z = 4

Now we use Identity

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

We know x+y+z=4

Therefore:

$x^3 + y^3 + z^3 - 3xyz = (4)(x^2+y^2+z^2-xy-yz-zx)$

Now it is given

xy + yz + zx = 5

And = - xy - yz - zx is same as -(xy + yz + zx)

Therefore:

$-xy-yz-zx=-5$

Simplified it is:

$x^3 + y^3 + z^3 - 3xyz = (4)(x^2+y^2+z^2-5)$

After that we use the first piece of info again

(x+y+z)^2=16

Using Identity

$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 16$

Now 2xy + 2yz + 2zx = 2(xy + yz + zx) which is just 2(5)=10

Therefore:

$x^2 + y^2 + z^2 + 10 = 16$

$x^2 + y^2 + z^2 = 16-10$

$x^2 + y^2 + z^2 = 6$

Now using previous equation

$x^3 + y^3 + z^3 - 3xyz = (4)(6-5)$

$x^3 + y^3 + z^3 - 3xyz = (4)(1)$

$x^3 + y^3 + z^3 - 3xyz = 4$