Question

Prove the following identities:
$\dfrac{{{a^3}\left( {b + c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{{b^3}\left( {c + a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{{a^3}\left( {a + b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}} = bc + ca + ab$

Answer 1

Hint: In this particular type of question use the concept that first convert the denominator of all the terms into standard format i.e. in (a – b), (b – c) and (c – a), then multiply and divide by appropriate values so that the denominator of all the terms become equal so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given equation:
$\dfrac{{{a^3}\left( {b + c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{{b^3}\left( {c + a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{{a^3}\left( {a + b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}} = bc + ca + ab$
Consider the L.H.S of the given equation we have,
$ \Rightarrow \dfrac{{{a^3}\left( {b + c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{{b^3}\left( {c + a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{{a^3}\left( {a + b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}}$
Now write denominator in the form of (a – b), (b – c) and (c – a) so we have,
$ \Rightarrow \dfrac{{{a^3}\left( {b + c} \right)}}{{ - \left( {a - b} \right)\left( {c - a} \right)}} + \dfrac{{{b^3}\left( {c + a} \right)}}{{ - \left( {b - c} \right)\left( {a - b} \right)}} + \dfrac{{{a^3}\left( {a + b} \right)}}{{ - \left( {c - a} \right)\left( {b - c} \right)}}$
Now in the above equation multiply and divide by (b – c) in the first term, (c – a) in the second term and (a – b) in the third term so we have,
$ \Rightarrow - \dfrac{{{a^3}\left( {b + c} \right)\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} - \dfrac{{{b^3}\left( {c + a} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} - \dfrac{{{a^3}\left( {a + b} \right)\left( {a - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
Now as we see that the denominator of all the terms is same so take the denominator term as a L.C.M and in numerator use the property that, $\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$ so we have,
$ \Rightarrow \dfrac{{ - {a^3}\left( {{b^2} - {c^2}} \right) - {b^3}\left( {{c^2} - {a^2}} \right) - {c^3}\left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
Now simplify the numerator of the above equation we have,
$ \Rightarrow \dfrac{{ - {a^3}{b^2} + {a^3}{c^2} - {b^3}{c^2} + {b^3}{a^2} - {c^3}{a^2} + {c^3}{b^2}}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
Now take $ - {a^2}{b^2}$ common from first and fourth term of the numerator, take ${c^2}$ from second and third term and $ - {c^3}$ from fifth and sixth term we have,
$ \Rightarrow \dfrac{{ - {a^2}{b^2}\left( {a - b} \right) + {c^2}\left( {{a^3} - {b^3}} \right) - {c^3}\left( {{a^2} - {b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
Now as we know that, $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$, and $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ so we have,
$ \Rightarrow \dfrac{{ - {a^2}{b^2}\left( {a - b} \right) + {c^2}\left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) - {c^3}\left( {a - b} \right)\left( {a + b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
$ \Rightarrow \left( {a - b} \right)\dfrac{{ - {a^2}{b^2} + {c^2}\left( {{a^2} + {b^2} + ab} \right) - {c^3}\left( {a + b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
$ \Rightarrow \dfrac{{ - {a^2}{b^2} + {c^2}\left( {{a^2} + {b^2} + ab} \right) - {c^3}\left( {a + b} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)}}$
Now again simplify the numerator of the above equation we have,
$ \Rightarrow \dfrac{{ - {a^2}{b^2} + {c^2}{a^2} + {c^2}{b^2} + {c^2}ab - a{c^3} - b{c^3}}}{{\left( {b - c} \right)\left( {c - a} \right)}}$
Now take $ - {a^2}$ common from first and second term of the numerator, take $b{c^2}$ from third and sixth term and $a{c^2}$ from fourth and fifth term we have,
$ \Rightarrow \dfrac{{ - {a^2}\left( {{b^2} - {c^2}} \right) + {c^2}b\left( {b - c} \right) + {c^2}a\left( {b - c} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)}}$
$ \Rightarrow \dfrac{{ - {a^2}\left( {b - c} \right)\left( {b + c} \right) + {c^2}b\left( {b - c} \right) + {c^2}a\left( {b - c} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)}}$
$ \Rightarrow \left( {b - c} \right)\dfrac{{ - {a^2}\left( {b + c} \right) + {c^2}b + {c^2}a}}{{\left( {b - c} \right)\left( {c - a} \right)}}$
$ \Rightarrow \dfrac{{ - {a^2}\left( {b + c} \right) + {c^2}b + {c^2}a}}{{\left( {c - a} \right)}}$
Now again simplify the numerator of the above equation we have,
$ \Rightarrow \dfrac{{ - {a^2}b - {a^2}c + {c^2}b + {c^2}a}}{{\left( {c - a} \right)}}$
Now take $\left( b \right)$ common from first and third term of the numerator, and take $ac$ from second and fourth term we have,
$ \Rightarrow \dfrac{{b\left( {{c^2} - {a^2}} \right) + ac\left( {c - a} \right)}}{{\left( {c - a} \right)}}$
$ \Rightarrow \dfrac{{b\left( {c - a} \right)\left( {c + a} \right) + ac\left( {c - a} \right)}}{{\left( {c - a} \right)}}$
$ \Rightarrow \left( {c - a} \right)\dfrac{{b\left( {c + a} \right) + ac}}{{\left( {c - a} \right)}}$
$ \Rightarrow b\left( {c + a} \right) + ac = bc + ba + ac$
= R.H.S
Hence Proved

Note: In such types of questions it is advised to simplify the LHS or the RHS according to their complexity of the given 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.