Question

Perform the indicated operation. (w 3 + 64) ÷ (4 + w)

Answer 1

Answer:

The expression $\frac{\left(w^3+64\right)}{w+4}$ becomes $w^2-4w+16$

Step-by-step explanation:

Given : Expression $\frac{\left(w^3+64\right)}{w+4}$

We have to find the simplified value of given expression.

Consider the given expression  $\frac{\left(w^3+64\right)}{w+4}$

Rewrite 64 as $4^3$

$=w^3+4^3$

$\mathrm{Apply\:Sum\:of\:Cubes\:Formula:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)$

$w^3+4^3=\left(w+4\right)\left(w^2-4w+4^2\right)$

Simplify,

$=\left(w+4\right)\left(w^2-4w+4^2\right)$

Given expression becomes,

$=\frac{\left(w+4\right)\left(w^2-4w+16\right)}{w+4}$

Cancel common factors, we have,

$=w^2-4w+16$

Thus, The expression $\frac{\left(w^3+64\right)}{w+4}$ becomes $w^2-4w+16$

Answer 2

W³ + 64 = w³ + 4³ = (w+4)³ - 3*w*4(w+4) = (w+4)[(w+4)² - 12w]

(w³ + 64) ÷ (w+4)
=(w+4)[(w+4)² - 12w] ÷ (w+4)
= (w+4)² - 12w
= w² -4w + 16