Decomposition factor (AX by) & sup3; + (by CZ) & sup3; - (AX CZ) & sup3;

Decomposition factor (AX by) & sup3; + (by CZ) & sup3; - (AX CZ) & sup3;

=（AX－by)³＋（by－cz）³+（cz-ax）³
Let AX = I by = k CZ = L
The above formula = (i-k) & sup3; + (K-L) & sup3; + (L-I) & sup3 ①
=3ik & sup2; - 3ki & sup2; + 3kl & sup2; - 3lk & sup2; + 3Li & sup2; - 3il & sup2; / / open the brackets
=3[(l-k)i²-(l²-k²)i+kl(l-k)]
=3(l-k)(i²-(l+k)i+kl)
=3(l-k)(i-k)(i-k)
Take AX = I by = k CZ = l into
So the original formula = 3 (CZ by) (AX by) (AX by)

a. B and C are the three sides of a triangle. If the system of equations ∑ x ^ 2-ax-y + B ^ 2 + AC = 0 ax-y + BC = 0 has only one real solution, what is the triangle
fast

Superposition, pin out y, get x ^ 2-2ax + B ^ 2 + AC BC = 0, and then take ERTA = 0, that is a-b-ac + BC = 0, using the complete square formula, get (a + B-C) (a-b) = 0, because a + b > C in the triangle, so a = B, that is isosceles triangle

The equation a (1 + I) x ^ 2 (1 + A ^ 2I) x + (a ^ 2 + I) = 0 has real roots

A (1 + I) x ^ 2 + (1 + A ^ 2I) x + (a ^ 2 + I) = 0. If so, separate the item with I from the item without I to get (AX ^ 2 + X + A ^ 2) + (AX ^ 2 + A ^ 2x + 1) I = 0. If a complex number equals 0, then its real part and imaginary part are equal to 0 respectively