Question

If $a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0,$ then $(a+b+c{)}^3$ = ?

• A. a + b + c = 0
• B. $(a+b+c{)}^3$=27abc
• C. a + b + c = 3abc
• D. $a^3+b^3+c^3$

Answer 1

Answer: (B)

$\because a^{\frac{1}{3}}+b^{\frac{1}{3}}+c^{\frac{1}{3}}=0\therefore {\left(a^{\frac{1}{3}}\right)}^3+{\left(b^{\frac{1}{3}}\right)}^3+{\left(c^{\frac{1}{3}}\right)}^3=3a^{\frac{1}{3}}{b}^{\frac{1}{3}}{c}^{\frac{1}{3}}\Rightarrow a+b+c=3a^{\frac{1}{3}}{b}^{\frac{1}{3}}{c}^{\frac{1}{3}}\Rightarrow (a+b+c{)}^3=(3a^{\frac{1}{3}}{b}^{\frac{1}{3}}{c}^{\frac{1}{3}}{)}^3\left(\text{Cubing both sides}\right)\Rightarrow (a+b+c{)}^3=27abc$