If $ a + b + c = 0 $ , then the value of $ {a^3} + {b^3} + {c^3} $ is equal to
(A) $ 0 $
(B) $ abc $
(C) $ 3abc $
(D) $ 2abc $
(i) $ \dfrac{1}{{\sqrt 5 }}{x^{\dfrac{1}{2}}} + 1 $ is a polynomial
(ii) $ \dfrac{{6\sqrt x + {x^{\dfrac{3}{2}}}}}{{\sqrt x }} $ is a polynomial, $ x \ne 0 $
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Ans: (C) is the correct option.
Here we know an identity,
$ {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $
It is given that, $ a + b + c = 0 $
Therefore, put it in the above equation
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $
$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 $
$ \Rightarrow {a^3} + {b^3} + {c^3} = 3abc $
Sample Question 1: Write whether the following statements are True or False.
Justify your answer.
Ans: It is a false statement.
Here, the exponent of a variable is not a whole number. So, it is not a polynomial.
Ans: It is a true statement.
We can also write it as,
$ \dfrac{{6\sqrt x + {x^{\dfrac{3}{2}}}}}{{\sqrt x }} = \dfrac{{\sqrt x {{\left( {6 + x} \right)}^{}}}}{{\sqrt x }} = 6 + x $
Here, the exponent of variable is not a whole number. So, it is a polynomial.