Question

If a + b + c = 0 then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$ is equal to:
(a) 2abc
(b) 3abc
(c) abc
(d) 4abc

Answer 1

Hint: Start by subtracting c on both the side in the given equation. There you will get the value of c. Then substitute the value of c in the expression. Then apply the formula of ${{\left( x+y \right)}^{3}}$ to reach a result.

Complete step-by-step answer:
Before starting with the solution to the above question, let us first discuss the important algebraic formulas that we are going to be using in our solution.
${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$ .
${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$ .
${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3{{y}^{2}}x$ .
${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ .

Now let us start with our solution to the above question.
In the question it is given that:
a + b + c = 0.
On taking c to the other side of the equation, we get
a + b = - c.

To get the value of c in terms of a and b, we will multiply both sides by -1. On doing so, we get
- ( a + b ) = c ………….(i)
Now we will substitute the value of c from equation (i) in the expression given in the question.
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$
$={{a}^{3}}+{{b}^{3}}+{{\left( -(a+b) \right)}^{3}}$
We know ${{\left( -x \right)}^{3}}=-{{x}^{3}}$ , applying this to our expression, we get
$={{a}^{3}}+{{b}^{3}}-{{\left( a+b \right)}^{3}}$

Now using the formula: ${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3{{y}^{2}}x$ , we get
$={{a}^{3}}+{{b}^{3}}-\left( {{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3{{b}^{2}}a \right)$
$={{a}^{3}}+{{b}^{3}}-{{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b-3{{b}^{2}}a$
$=-3{{a}^{2}}b-3{{b}^{2}}a$

Now to simplify the expression let us take -3ab common from both terms, then our expression becomes:
 $=-3ab(a+b)$
Again substituting the value of c from equation (i), our expression simplifies to:
$=-3ab(-c)$
= 3abc.
So, we can say that the expression ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}$ is equal to 3abc provided a + b + c = 0.

Therefore, our final answer is option (b) 3abc.

Note: While opening the brackets, keep a close check on the signs, as in such questions, the signs play a crucial role. The general mistake a student makes is : $-b-c=-\left( b-c \right)$ . It is preferred to keep the expression as neat as possible by eliminating all the removable terms as the larger is the number of terms in your expression the more is the probability of making a mistake.
For solving the above question, we can also use the formula: ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$ .