Let's expand: ${(x-3)}^2$

ⓘ Raising a sum to a positive integer power means multiplying it by itself as many times as the exponent indicates: ${(x-3)}^2$ = $(x-3)·(x-3)$

$(x-3)·(x-3)$

ⓘ Expand using the distributive law, multiplying each term of the sum by the other factor: $(x-3)·(x-3)$ = $x·x$$+x·\left(-3\right)$$-3x$$-3·\left(-3\right)$

$x·x$$+x·\left(-3\right)$$-3x$$-3·\left(-3\right)$

ⓘ To multiply powers with the same base, add their exponents: $x·x$ = $x^2$■Write the minus sign in front of the product: $x·\left(-3\right)$ = $-3x$■Perform multiplications before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: $-3·\left(-3\right)$ = $9$

$x^2$$-3x$$-3x$$+9$

ⓘ Group like terms by adding their coefficients: $-3x-3x$ = $-6x$

▶Expanded form:   $x^2-6x+9$