Question

Show that \[{a^2}{b^2}\left( {a - b} \right) + {b^2}{c^2}\left( {b - c} \right) + {c^2}{a^2}\left( {c - a} \right)\] us exactly divisible by \[a - b\], \[b - c\] and \[c - a\]?

Answer 1

Hint: The given equation consists of three variables a, b and c. Hence, to prove the given expression \[{a^2}{b^2}\left( {a - b} \right) + {b^2}{c^2}\left( {b - c} \right) + {c^2}{a^2}\left( {c - a} \right)\] which is exactly divisible by \[a - b\], \[b - c\] and \[c - a\]. We know that the algebraic expression should be any one of the forms such as addition, subtraction, multiplication and division, hence in the given equation there is a constant variable involved and to get this equation, combine all the like terms and then simplify the terms to get the divisibility of the required terms.

Complete step by step answer:
Given,
\[{a^2}{b^2}\left( {a - b} \right) + {b^2}{c^2}\left( {b - c} \right) + {c^2}{a^2}\left( {c - a} \right)\] …………….. 1
Let us keep the first term \[{a^2}{b^2}\left( {a - b} \right)\] constant and simplify the other terms by multiplying with their respective terms i.e.,
\[ = {a^2}{b^2}\left( {a - b} \right) + {b^3}{c^2} - {b^2}{c^3} + {c^3}{a^2} - {c^2}{a^3}\] ……………….. 2
As, we can see that variable c is common in equation 2, hence we get:
\[ = {a^2}{b^2}\left( {a - b} \right) - {c^2}\left( {{a^3} - {b^3}} \right) + {c^3}\left( {{a^2} - {b^2}} \right)\]
As, we need to show that equation is exactly divisible by \[a - b\], \[b - c\] and \[c - a\], hence let us keep the first term \[a - b\] constant, and hence multiply the other terms as:
\[ = \left( {a - b} \right)\left( {{a^2}{b^2} - {c^2}\left( {{a^2} + ab + {b^2}} \right) + {c^3}\left( {a + b} \right)} \right)\]
Multiplying the terms, we get:
\[ = \left( {a - b} \right)\left( {{a^2}{b^2} - {c^2}{a^2} - {c^2}ab - {c^2}{b^2} + {c^3}a + {c^3}b} \right)\]
As the next term to show the divisibility is \[b - c\], hence simplify the common terms with respect to it as:
\[ = \left( {a - b} \right)\left( {{a^2}\left( {{b^2} - {c^2}} \right) - {c^2}a\left( {b - c} \right) - {c^2}b\left( {b - c} \right)} \right)\]
\[ = \left( {a - b} \right)\left( {b - c} \right)\left( {{a^2}b + {a^2}c - {c^2}a - {c^2}b} \right)\]
Now, the remaining terms is \[c - a\], hence simplify the common terms with respect to it as:
\[ = \left( {a - b} \right)\left( {b - c} \right)\left( {b\left( {{a^2} - {c^2}} \right) + ac\left( {a - c} \right)} \right)\]
Hence, we get the required form as:
\[ = \left( {a - b} \right)\left( {b - c} \right)\left( {a - c} \right)\left( {ab + bc + ca} \right)\]
This factorisation proves that the given expression is exactly divisible by \[a - b\], \[b - c\] and \[c - a\].

Note: The key point is that, when solving math equations, we must always keep the equation balanced so that both sides are always equal and we must know that a two-step equation is an algebraic equation that takes you two steps to solve. You've solved the equation when you get the variable by itself, with no numbers in front of it, on one side of the equal sign. Suppose you want to prove a theorem in the form "For all integers n greater than equal to a, \[P\left( n \right)\] is true". \[P\left( n \right)\] must be an assertion that we wish to be true for all \[n = a,a + 1,....\] ; like a formula.