From the first equation:
$4(x^2 + y^2 + z^2) = a$
Now, substitute this value of $a$ into the second equation:
$4(xyz) = 3 + a = 3 + 4(x^2 + y^2 + z^2)$
Simplifying:
$4(xyz) = 3 + 4(x^2 + y^2 + z^2)$
Let's assume $x = y = z$, so the equations become:
$4(3x^2) = a$ and $4x^3 = 3 + a$
From the first equation:
$12x^2 = a$
Substitute this into the second equation:
$4x^3 = 3 + 12x^2$
Solving for $x$:
$x^3 = \frac{3 + 12x^2}{4}$
By trial, we find $x = 1$ satisfies both equations, so:
$a = 4(1^2 + 1^2 + 1^2) = 12$
Thus, $a = 3$.
If the set of all values of \( a \), for which the equation \( 5x^3 - 15x - a = 0 \) has three distinct real roots, is the interval \( (\alpha, \beta) \), then \( \beta - 2\alpha \) is equal to
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .