Respuesta :

Answer:

Yes.

Step-by-step explanation:

Yes, because we can replace x^3 by another variable, say y:

x^9 - 5x^3 + 6 = 0  

Let x^3 = y, then  we have x^9 = (x^3)^2 = y^2:

y^2 - 5y + 6 = 0  which is quadratic from.

Answer:

The equation is not quadratic in form.

The variable part of the first term is not the square of the variable part of the

second term.

The result of squaring x cubed is x to the 6th power, not the 9th power.

x9(not equal to) (x^3)^2

Step-by-step explanation: