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mskyeg
 2 years ago
What are the real or imaginary solutions of the polynomial equation?
x3 = 216
mskyeg
 2 years ago
What are the real or imaginary solutions of the polynomial equation? x3 = 216

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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(x^3216=0 \\ x^36^3=0 \\ \text{now apply }a^3b^3 formula\)

mskyeg
 2 years ago
Best ResponseYou've already chosen the best response.0like what do i plug intohe formula

mskyeg
 2 years ago
Best ResponseYou've already chosen the best response.0how does that help all i get is x^3  216

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1thats because u don't know that formula

mskyeg
 2 years ago
Best ResponseYou've already chosen the best response.0oooh i thought that WAS the formula

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\large a^3b^3=(ab)(a^2+ab+b^2)\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\((x6)(x^2+6x+36)=0 \\ (x6)=0,(x^2+6x+36)=0 \\ x=6,\text{use quadratic formula for x^2+6x+36=0}\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1Compare your quadratic equation with \(ax^2+bx+c=0\) find a,b,c then the two roots of x are: \(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\)

mskyeg
 2 years ago
Best ResponseYou've already chosen the best response.0I tried but i got a really weird answer

mskyeg
 2 years ago
Best ResponseYou've already chosen the best response.0a=6 b=3+i square root 54 c=3i square root 54

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1so two other values of x are ?

mskyeg
 2 years ago
Best ResponseYou've already chosen the best response.0what about the other two

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1thats what u have u work out using \(\huge{x=\frac{b \pm \sqrt{b^24ac}}{2a}}\) u have a,b,c values correct.
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