Question

Prove that: ${{\left( a+b+c \right)}^{3}}-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=3\left( a+b \right)\left( b+c \right)\left( c+a \right)$.

Answer 1

Hint: Start with the left-hand side by applying the formula of ${{x}^{3}}-{{y}^{3}}$ followed by factorisation at the end to reach the required expression.

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.
${{x}^{3}}-{{y}^{3}}=\left( x-y \right)\left( {{x}^{2}}+xy+{{y}^{2}} \right)$ .
${{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)$ .
${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx$ .

Now let us start with our solution with the left-hand side of the equation that is given in the question.
${{\left( a+b+c \right)}^{3}}-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}$
$=\left( {{\left( a+b+c \right)}^{3}}-{{a}^{3}} \right)-\left( {{b}^{3}}+{{c}^{3}} \right)$

Now applying the formula of ${{x}^{3}}-{{y}^{3}}$ and ${{x}^{3}}+{{y}^{3}}$ to our expression, we get
$\left( \left( a+b+c \right)-a \right)\left( {{(a+b+c)}^{2}}+a\left( a+b+c \right)+{{a}^{2}} \right)-\left( b+c \right)\left( {{b}^{2}}-bc+{{c}^{2}} \right)$
$=\left( b+c \right)\left( {{(a+b+c)}^{2}}+{{a}^{2}}+ab+ac+{{a}^{2}} \right)-\left( b+c \right)\left( {{b}^{2}}-bc+{{c}^{2}} \right)$
Taking ( b + c ) common from each term of the expression, we get
$\left( b+c \right)\left( \left( {{(a+b+c)}^{2}}+{{a}^{2}}+ab+ac+{{a}^{2}} \right)-\left( {{b}^{2}}-bc+{{c}^{2}} \right) \right)$

Now applying the formula of ${{\left( x+y+z \right)}^{2}}$ , we get
$\left( b+c \right)\left( \left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca+{{a}^{2}}+ab+ac+{{a}^{2}} \right)-\left( {{b}^{2}}-bc+{{c}^{2}} \right) \right)$
\[=\left( b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca+{{a}^{2}}+ab+ac+{{a}^{2}}-{{b}^{2}}+bc-{{c}^{2}} \right)\]
\[\Rightarrow \left( b+c \right)\left( 3{{a}^{2}}+3ab+3bc+3ca \right)\]
Taking 3 common, we get
\[3\left( b+c \right)\left( {{a}^{2}}+ab+bc+ca \right)\]
\[=3\left( b+c \right)\left( a\left( a+b \right)+c\left( b+a \right) \right)\]
\[=3\left( a+b \right)\left( b+c \right)\left( c+a \right)\]
So, we can say that left-hand side = right-hand side = 3 ( a + b )( b + c )( c + a) .

Hence, we have proved that ${{\left( a+b+c \right)}^{3}}-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=3\left( a+b \right)\left( b+c \right)\left( c+a \right)$ .

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.
In the above question we could have used the formula of ${{\left( a+b+c \right)}^{3}}$ , but this might give you 10 terms making it difficult to factorise to the form given on the other side of the equation.