Respuesta :
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
(w³ + 64) ÷ (w+4)
=(w+4)[(w+4)² - 12w] ÷ (w+4)
= (w+4)² - 12w
= w² -4w + 16
Answer:
The expression [tex]\frac{\left(w^3+64\right)}{w+4}[/tex] becomes [tex]w^2-4w+16[/tex]
Step-by-step explanation:
Given : Expression [tex]\frac{\left(w^3+64\right)}{w+4}[/tex]
We have to find the simplified value of given expression.
Consider the given expression [tex]\frac{\left(w^3+64\right)}{w+4}[/tex]
Rewrite 64 as [tex]4^3[/tex]
[tex]=w^3+4^3[/tex]
[tex]\mathrm{Apply\:Sum\:of\:Cubes\:Formula:\:}x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)[/tex]
[tex]w^3+4^3=\left(w+4\right)\left(w^2-4w+4^2\right)[/tex]
Simplify,
[tex]=\left(w+4\right)\left(w^2-4w+4^2\right)[/tex]
Given expression becomes,
[tex]=\frac{\left(w+4\right)\left(w^2-4w+16\right)}{w+4}[/tex]
Cancel common factors, we have,
[tex]=w^2-4w+16[/tex]
Thus, The expression [tex]\frac{\left(w^3+64\right)}{w+4}[/tex] becomes [tex]w^2-4w+16[/tex]
