Perform the indicated operation. (w 3 + 64) ÷ (4 + w)
w³ + 64 = w³ + 4³ = (w+4)³ - 3w4(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
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 \(w^2-4w+16\)