Answer
$(x-\dfrac{1}{6})^2$
Work Step by Step
Given: $x^2-\dfrac{1}{3}x$
We need to add $(\dfrac{-1}{6})^2$ to complete the square.
Thus,
$x^2-\dfrac{1}{3}x+(\dfrac{-1}{6})^2=x^2+\dfrac{1}{36}-\dfrac{1}{3}x$
Apply difference of polynomial formula.
or, $x^2+(\dfrac{1}{6})^2-2(x)(\dfrac{1}{6})=(x-\dfrac{1}{6})^2$
