y = x (x – 3)² decreases for the values of x given by

The correct option is 1 < x < 3.

To find the values of $x$ for which the function $y=x{(x–3)}^2$ decreases, we need to analyze its derivative. A function decreases when its first derivative with respect to $x$ is negative, i.e., $\frac{dy}{dx}<0$.

First, let's expand the function to make differentiation simpler:
$y=x{(x–3)}^2$
Using the algebraic identity ${(a–b)}^2=a^2–2ab+b^2$, we have:
${(x–3)}^2=x^2–6x+9$
Now, multiplying by $x$:
$y=x(x^2–6x+9)=x^3–6x^2+9x$

Next, we find the first derivative of $y$ with respect to $x$:
$\frac{dy}{dx}=\frac{d}{dx}(x^3–6x^2+9x)$
Applying the power rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n–1}$:
$\frac{dy}{dx}=3x^2–12x+9$

To find the interval where the function decreases, set $\frac{dy}{dx}<0$:
$3x^2–12x+9<0$
Divide the entire inequality by 3 to simplify:
$x^2–4x+3<0$

Factor the quadratic expression by finding two numbers that multiply to 3 and add up to -4 (which are -1 and -3):
$(x–1)(x–3)<0$

For the product of two terms to be negative, one factor must be positive and the other must be negative. This occurs when the value of $x$ lies strictly between the roots of the quadratic equation, which are $x=1$ and $x=3$.
Therefore, the inequality holds true when:
$1<x<3$

Hence, the function decreases for the values of $x$ given by 1 < x < 3.