Question

X+y+z=x^2+y^2+z^2=x^3+y^3+z^3=12 x^4+y^4+z^4=?

Answer 1

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Answer 2

To find the value of $x^4 + y^4 + z^4$, we can use the given information and the identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz)$.

Given: $x + y + z = 12$ $x^2 + y^2 + z^2 = 12$ $x^3 + y^3 + z^3 = 12$

Using the identity above, we can rewrite the equation for $x^3 + y^3 + z^3$: $x^3 + y^3 + z^3 - 3xyz = 12 - 3xyz$

Since we know that $x^3 + y^3 + z^3 = 12$, we can substitute this value in the equation above: $12 - 3xyz = 12$

Simplifying the equation, we get: $3xyz = 0$

This equation implies that at least one of the variables $x$, $y$, or $z$ must be zero.

Now let's consider the equation $x^2 + y^2 + z^2 = 12$. Since one of the variables must be zero, let's assume $z = 0$.

Substituting $z = 0$ into the equation $x + y + z = 12$, we get: $x + y = 12$

Squaring this equation, we get: $x^2 + 2xy + y^2 = 144$

Since $x^2 + y^2 = 12$ (from the given information), we can rewrite the equation as: $12 + 2xy = 144$

Simplifying, we find: $xy = 66$

Now let's calculate the value of $x^4 + y^4 + z^4$ using the given information.

Using the equation $x^2 + y^2 + z^2 = 12$ and substituting $z = 0$, we have: $x^2 + y^2 = 12$

Squaring this equation, we get: $x^4 + 2x^2y^2 + y^4 = 144$

Since $x^2y^2 = (xy)^2$ and we know that $xy = 66$, we can rewrite the equation as: $x^4 + 2(66)^2 + y^4 = 144$

Simplifying, we find: $x^4 + 4356 + y^4 = 144$

Rearranging the equation, we get: $x^4 + y^4 = 144 - 4356$

Finally, solving the equation, we find: $x^4 + y^4 = -4212$

Therefore, $x^4 + y^4 + z^4 = -4212$