Question

Ajutor S 5 va rog mult

Answer 1

$\left|\begin{array}{ccc}x+1&1&2\\x-1&x&-3\\0&x&1-x\end{array}\right| = \left|\begin{array}{cc}x&1\\-1&5\end{array}\right|\\\\x(x+1)(1-x) + 2x(x-1) + 3x(x+1) - (x-1)(1-x) = 5x + 1\\x(1-x^2) + 2x^2 - 2x + 3x^2 + 3x + (x-1)^2 = 5x + 1\\-x^3 + x + 5x^2 + x + x^2 - 2x + 1 = 5x + 1\\-x^3 + 6x^2 - 5x = 0\\-x(x^2 - 6x + 5) = 0\\-x(x^2 - x - 5x + 5) = 0\\-x[x(x-1)-5(x-1)] = 0\Rightarrow -x(x-1)(x-5) = 0\\\Rightarrow x = 0 \text{ sau } x = 1 \text{ sau } x = 5\Rightarrow x_1 = 0,\, x_2 = 1,\, x_3 = 5$

$S = x_1^3 + x_2^3 + x_3^3 = 0^3 + 1^3 + 5^3 = 126$