Question

If $ x + y + z = 0, $ then show that $ {x^3} + {y^3} + {z^3} = 3xyz $

Answer 1

Hint: Use the expansion formula of $ {(x + y)^3} $ . Then replace $ y $ with $ y + z $ to write it in three variables. Further expand it using the condition given in the question and simplify it in the terms of the required result.

Complete step-by-step answer:
It is given in the question that
 $ x + y + z = 0 $ . . . . (1)
We have to prove $ {x^3} + {y^3} + {z^3} = 3xyz $
We know that,
 $ {(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) $ . . . (2)
By rearranging it, we can write
 $ {a^3} + {b^3} = {(a + b)^3} - 3ab(a + b) $
Now, replace, $ a $ with $ x $ and $ b $ with $ y + z $ . Then we get
 $ {x^3} + {(x + y)^3} = {(x + y + z)^3} - 3x(y + z)(x + y + z) $
By using equation (1) we can simplify the RHS of the above equation. And by using equation (2), we can further expand LHS of the above equation. Hence, by using equation (1) and (2), we get
 $ {x^3} + {y^3} + {z^3} + 3yz(y + z) = 0 - 0 $ . . . (3)
Now, since,
 $ x + y + z = 0 $
We get
 $ y + z = - x $
By substituting this into equation (3), we get
 $ {x^3} + {y^3} + {z^3} + 3yz( - x) = 0 $
 $ \Rightarrow {x^3} + {y^3} + {z^3} - 3yzx = 0 $
By rearranging it, we can write
 $ {x^3} + {y^3} + {z^3} = 3xyz $
Hence proved.

Note: To solve this question, you need to know the expansion formula of cubic form. And you need to understand that the condition given in the question is useful, and that is why it is given in the question. So you need to formulate your solution in such a way that you could get to use the condition given in the question. This approach always helps to get to the result fast and without errors.
Equation (2) is the result of the following formula
 $ {(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} $
Which can be rearranged as
 $ {(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2} $
Now, by taking $ 3ab $ common, we can further write it as
 $ {(a + b)^3} = {a^3} + {b^3} + 3ab(a + b) $
We used this result to solve the question.