Question

Factorise \[ab\left( {{x^2} + {y^2}} \right) - xy\left( {{a^2} + {b^2}} \right)\].

Answer 1

Hint: In order to factorise the expression \[ab\left( {{x^2} + {y^2}} \right) - xy\left( {{a^2} + {b^2}} \right)\], we will first simplify the given expression by multiplying the common terms with the terms combined in the brackets. Then we will rearrange the obtained expression and then we will take terms common in such a way that it gets simplified.

Complete step by step answer:
Given expression is \[ab\left( {{x^2} + {y^2}} \right) - xy\left( {{a^2} + {b^2}} \right)\].
On multiplying the common terms with the terms combined in the brackets, we get
\[ = \left( {ab{x^2} + ab{y^2}} \right) - \left( {xy{a^2} + xy{b^2}} \right)\]
On simplifying, we get
\[ = ab{x^2} + ab{y^2} - xy{a^2} - xy{b^2}\]
On rearranging we get,
\[ = ab{x^2} - xy{a^2} + ab{y^2} - xy{b^2}\]
Taking common factors out and combining rest of the expression in the brackets we get
\[ = ax\left( {bx - ay} \right) + by\left( {ay - bx} \right)\]
On rewriting we get
\[ = ax\left( {bx - ay} \right) - by\left( {bx - ay} \right)\]
Taking \[\left( {bx - ay} \right)\] common from the above expression, we get
\[ = \left( {ax - by} \right)\left( {bx - ay} \right)\]
Therefore, factorisation of \[ab\left( {{x^2} + {y^2}} \right) - xy\left( {{a^2} + {b^2}} \right)\] is \[\left( {ax - by} \right)\left( {bx - ay} \right)\].

Note:
Some methods of factorisation are:
\[1\]. By the method of common factors in which each term of given algebraic expression is written as a product of irreducible factors and the common factors are taken out and the rest of the expression is combined in the brackets.
\[2\]. By regrouping, terms can be grouped in such a way that all the terms in each group have a common factor. A common factor comes out from all the groups and leads to required factorisation of the expression.
\[3\]. By using different identities.
\[4\]. By factorising an algebraic expression of the form \[{x^2} + px + q\], by finding two factors \[a\] and \[b\] such that \[ab = q\] and \[a + b = p\].