x³+64=0
If we find first zeroes of this cubic polynomial equation we have to replace X by negative 4
Since it is cubic equation so it has three roots
Imply -4 is the one root of this equation
Now X³= -64
Imply X = ± (-64)⅓
Imply X = ± i {(4)³}⅓
Imply X = 4i , – 4i
Thus the required roots of given equation are as :
4i , – 4i, -4
-4
-4
-4
-4
-4
-4
-4,-4w,-4w^2
-4,-4w,-4w^2
x=-4
X=-4
X1=-4, X2=2+(3.46410161514)i, X3=2-(3.46410161514)i
-4
-4
X=-4
-4
X=-4
-4
-4
X☞ -4
(x+4)(x^2-4x+16)=(x+4)(x-8)(x+4) sum between different cubes , third degree
-4×-4= 16
-4×-4×-4= -64
X³+64=0
X³=-64
X= -4
x³+64=0
If we find first zeroes of this cubic polynomial equation we have to replace X by negative 4
Since it is cubic equation so it has three roots
Imply -4 is the one root of this equation
Now X³= -64
Imply X = ± (-64)⅓
Imply X = ± i {(4)³}⅓
Imply X = 4i , – 4i
Thus the required roots of given equation are as :
4i , – 4i, -4
X³+64=0
X³= _64
X = _ 4
X= – 4 or 4i
-4:::::-4w:::::::-4w^2
-4
_4
4i
x=-4
X= -4
X=-4
3 soutions
X=-4