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