Question

Prove that:
$\dfrac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}} = 1$

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:
$\dfrac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}} = 1$
Proof –
Consider the L.H.S part of the above equation we have,
$ \Rightarrow \dfrac{{\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{\left( {x - a} \right)\left( {x - 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{{\left( {x - b} \right)\left( {x - c} \right)}}{{ - \left( {a - b} \right)\left( {c - a} \right)}} + \dfrac{{\left( {x - c} \right)\left( {x - a} \right)}}{{ - \left( {b - c} \right)\left( {a - b} \right)}} + \dfrac{{\left( {x - a} \right)\left( {x - 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{{\left( {b - c} \right)\left( {x - b} \right)\left( {x - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} - \dfrac{{\left( {c - a} \right)\left( {x - c} \right)\left( {x - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} - \dfrac{{\left( {a - b} \right)\left( {x - a} \right)\left( {x - b} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
Now simplify the numerator we have,
$ \Rightarrow - \dfrac{{\left( {b - c} \right)\left( {{x^2} - bx - cx + bc} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} - \dfrac{{\left( {c - a} \right)\left( {{x^2} - cx - ax + ac} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} - \dfrac{{\left( {a - b} \right)\left( {{x^2} - ax - bx + ab} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$
\[
   \Rightarrow - \dfrac{{\left( {{x^2}b - {b^2}x - bcx + {b^2}c - {x^2}c + bcx + {c^2}x - b{c^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \\
   - \dfrac{{\left( {{x^2}c - {c^2}x - acx + a{c^2} - {x^2}a + acx + {a^2}x - {a^2}c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \\
   - \dfrac{{\left( {{x^2}a - {a^2}x - abx + {a^2}b - {x^2}b + abx + {b^2}x - a{b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \\
\]
Now cancel out the common terms from the numerator when we take the L.C.M we have,
\[ \Rightarrow \dfrac{{ - \left( {{b^2}c - b{c^2}} \right) - \left( {a{c^2} - {a^2}c} \right) - \left( {{a^2}b - a{b^2}} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
\[ \Rightarrow \dfrac{{ - {b^2}c + b{c^2} - a{c^2} + {a^2}c - {a^2}b + a{b^2}}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\]
Now expand the denominator we have,
\[ \Rightarrow \dfrac{{ - {b^2}c + b{c^2} - a{c^2} + {a^2}c - {a^2}b + a{b^2}}}{{\left( {a - b} \right)\left( {bc - ba - {c^2} + ac} \right)}}\]
\[ \Rightarrow \dfrac{{ - {b^2}c + b{c^2} - a{c^2} + {a^2}c - {a^2}b + a{b^2}}}{{\left( {abc - b{a^2} - a{c^2} + {a^2}c - {b^2}c + a{b^2} + b{c^2} - abc} \right)}}\]
\[ \Rightarrow \dfrac{{ - {b^2}c + b{c^2} - a{c^2} + {a^2}c - {a^2}b + a{b^2}}}{{\left( { - b{a^2} - a{c^2} + {a^2}c - {b^2}c + a{b^2} + b{c^2}} \right)}}\]
Now arrange the denominator according to the numerator we have,
\[ \Rightarrow \dfrac{{ - {b^2}c + b{c^2} - a{c^2} + {a^2}c - {a^2}b + a{b^2}}}{{\left( { - {b^2}c + b{c^2} - a{c^2} + {a^2}c - {a^2}b + a{b^2}} \right)}}\]
So as we see that numerator and denominator is same so it all cancel out, so we have,
 = 1
= 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.