Question : If $\sqrt[3]{a}+\sqrt[3]{b}=\sqrt[3]{c}$, then the simplest value of $\left (a+b+c \right )^{3}+27abc$ is:
Option 1: $-1$
Option 2: $3$
Option 3: $-3$
Option 4: $0$
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Correct Answer: $0$
Solution : Given: $\sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{c}$ ⇒ $\sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c} = 0$ We know that if $x+y+z=0$, $ x^3+y^3+z^3=3xyz$ $\therefore \ a + b - c = -3\left(abc\right)^{\frac{1}{3}}$ On cubing both sides, ⇒ $(a+b-c)^3=-27abc$ $\therefore (a + b - c)^3 + 27abc=0$ Hence, the correct answer is $0$.
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