If $a^2 + b^2 + c^2 + 48 = 8(a + b + c)$, then what is the value of \(\sqrt[3]{a^3-b^3+c^3}\) ?

4

a² + b² + c² + 48 = 8(a + b + c)

= a² - 8a + 16 + b² - 8b + 16 + c² - 8c + 16 = 0

= (a - 4)² + (b - 4)² + (c - 4)²  = 0

When the sum of the squares is 0 then the value of each of them is equal to 0.

So,

 a - 4 = 0

b - 4 = 0

c - 4 = 0

= a = b = c = 4e

Now the value of \(\sqrt[3]{a^3-b^3+c^3}\)  = \(\sqrt[3]{4^3-4^3+4^3}\) = \(\sqrt[3]{64}\)  = 4