Question

the dimensions of a rectangle are shown. write the area of the rectangle as a sum of cubes. the area of the rectangle as a sum of cubes is (square). (x^{2}-3x + 9) (x + 3)

Explanation:

Step1: Recall area formula for rectangle

The area $A$ of a rectangle is given by $A = \text{length}\times\text{width}$. Here, length $l=x + 3$ and width $w=x^{2}-3x + 9$.

Step2: Multiply the two polynomials

$(x + 3)(x^{2}-3x + 9)=x(x^{2}-3x + 9)+3(x^{2}-3x + 9)$.

Step3: Distribute $x$ and 3

$x(x^{2}-3x + 9)=x^{3}-3x^{2}+9x$ and $3(x^{2}-3x + 9)=3x^{2}-9x + 27$.

Step4: Combine like - terms

$(x^{3}-3x^{2}+9x)+(3x^{2}-9x + 27)=x^{3}+(-3x^{2}+3x^{2})+(9x-9x)+27$.
$=x^{3}+0 + 0+27=x^{3}+27$.

Answer:

$x^{3}+27$