Question

Prove the following identity:
\[\sum {{a^2}\left( {b + c} \right) - \sum {{a^3}} - 2abc} = \left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)\]

Answer 1

Hint: In this particular question use the concept of simplification by adding and subtracting the terms such that it will give us the required answer, so use this concept to reach the solution of the question.

Complete step-by-step solution:
Given equation:
\[\sum {{a^2}\left( {b + c} \right) - \sum {{a^3}} - 2abc} = \left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)\]
Consider the LHS part of the above equation we have,
\[ \Rightarrow \sum {{a^2}\left( {b + c} \right) - \sum {{a^3}} - 2abc} \]
Now expand the summation we have,
\[ \Rightarrow {a^2}\left( {b + c} \right) + {b^2}\left( {c + a} \right) + {c^2}\left( {a + b} \right) - {a^3} - {b^3} - {c^3} - 2abc\]
Now simplify it we have,
\[ \Rightarrow {a^2}b + {a^2}c + {b^2}c + {b^2}a + {c^2}a + {c^2}b - {a^3} - {b^3} - {c^3} - 2abc\]
Now arrange the terms we have,
\[ \Rightarrow \left( { - {a^3} + {b^2}a + {c^2}a - 2abc} \right) + \left( { - {b^3} + {c^2}b + {a^2}b} \right) + \left( { - {c^3} + {a^2}c + {b^2}c} \right)\]
Now take a, b and c common from first, second and third terms respectively we have,
\[ \Rightarrow a\left( { - {a^2} + {b^2} + {c^2} - 2bc} \right) + b\left( { - {b^2} + {c^2} + {a^2}} \right) + c\left( { - {c^2} + {a^2} + {b^2}} \right)\]
Now add by $\left( { - 2b{c^2} + 2{b^2}c} \right)$ in second term and subtract by $\left( { - 2b{c^2} + 2{b^2}c} \right)$in third term respectively so that overall no change we have,
\[ \Rightarrow a\left( { - {a^2} + {b^2} + {c^2} - 2bc} \right) + \left[ {b\left( { - {b^2} + {c^2} + {a^2}} \right) - 2b{c^2} + 2{b^2}c} \right] + \left[ {c\left( { - {c^2} + {a^2} + {b^2}} \right) - \left( { - 2b{c^2} + 2{b^2}c} \right)} \right]\]Now take b and c common from second and third term of the above equation respectively we have,
\[ \Rightarrow a\left( { - {a^2} + {b^2} + {c^2} - 2bc} \right) + b\left[ {\left( { - {b^2} + {c^2} + {a^2}} \right) - 2{c^2} + 2bc} \right] + c\left[ {\left( { - {c^2} + {a^2} + {b^2}} \right) - \left( { - 2bc + 2{b^2}} \right)} \right]\]
Now simplify the above equation we have,
\[ \Rightarrow - a\left( {{a^2} - {b^2} - {c^2} + 2bc} \right) + b\left[ {{a^2} - {b^2} - {c^2} + 2bc} \right] + c\left[ {{a^2} - {b^2} - {c^2} + 2bc} \right]\]
\[ \Rightarrow \left( { - a + b + c} \right)\left( {{a^2} - {b^2} - {c^2} + 2bc} \right)\]................... (1)
Now simplify the second term of the above equation we have,
\[ \Rightarrow \left( {{a^2} - {b^2} - {c^2} + 2bc} \right)\]
Now add and subtract by ac and ab in the above equation we have,
\[ \Rightarrow \left( {{a^2} - {b^2} - {c^2} + 2bc} \right) = \left( {{a^2} - {b^2} - {c^2} + 2bc + ac + ab - ac - ab} \right)\]
Now rearranging the terms of the above equation we have,
\[ \Rightarrow \left( {{a^2} - {b^2} - {c^2} + 2bc} \right) = \left( { - {c^2} + ac + bc} \right) + \left( {{a^2} + ab - ac} \right) + \left( { - {b^2} - ab + bc} \right)\]
Now take c, a and –b common from the above terms respectively we have,
\[ \Rightarrow \left( {{a^2} - {b^2} - {c^2} + 2bc} \right) = c\left( { - c + a + b} \right) + a\left( {a + b - c} \right) - b\left( {b + a - c} \right)\]
\[ \Rightarrow \left( {{a^2} - {b^2} - {c^2} + 2bc} \right) = \left( { - c + a + b} \right)\left( {c + a - b} \right)\]
Now substitute this value in equation (1) we have,
\[ \Rightarrow \left( { - a + b + c} \right)\left( { - c + a + b} \right)\left( {c + a - b} \right)\]
\[ \Rightarrow \left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)\]
= RHS
Hence Proved

Note: In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of the functions. Sometimes proving LHS = RHS needs simplification on both sides of the equation. Remember to convert dissimilar functions to get to the final result, and check whether R.H.S is equal to L.H.S or not if yes then it is the required answer.