Question

(x+3)(x^2-3x+9)+x(3-x)(3+x)=0

Answer 1

$(x+3)(x^2-3x+9)+x(3-x)(3+x)=0\\(x+3)(x^2-3x+9)+x(3-x)(x+3)=0\\(x+3)((x^2-3x+9)+x(3-x))=0\\(x+3)(x^2-3x+9+3x-x^2)=0\\(x+3)*9=0\\x+3=0\\x=-3$

Answer 2

To solve the equation, let's first simplify and expand the terms:

(x+3)(x^2-3x+9) + x(3-x)(3+x) = 0

Expanding the first part (x+3)(x^2-3x+9):

(x+3)(x^2-3x+9) = x(x^2-3x+9) + 3(x^2-3x+9) = x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 = x^3 + 27

Expanding the second part x(3-x)(3+x):

x(3-x)(3+x) = x(3^2 - x^2) = x(9 - x^2) = 9x - x^3

Now, let's rewrite the equation:

x^3 + 27 + 9x - x^3 = 0

The x^3 and -x^3 terms cancel out, leaving us with:

27 + 9x = 0

Now, isolate the term with x:

9x = -27

Finally, divide both sides by 9:

x = -27/9

Simplify the fraction:

x = -3

Therefore, the solution to the equation is x = -3.