Question

Find the HCF of the following triples:
$a^{4}b-a{b}^4,a^4{b}^2-a^2{b}^4$ and $a^2{b}^2(a^4-b^4)$

Answer 1

We know that $HCF$ is the highest common factor.

Factorise $a^{4}b-a{b}^4$ as follows:

$a^{4}b-a{b}^4=ab(a^3-b^3)=ab(a-b)(a^2+b^2+ab)$

(using identity $a^3-b^3=(a-b)(a^2+b^2+ab)$)

Now, factorise $a^4{b}^2-a^2{b}^4$ as follows:

$a^4{b}^2-a^2{b}^4=a^2{b}^2(a^2-b^2)=a^2{b}^2(a+b)(a-b)$ (using identity $a^2-b^2=(a+b)(a-b)$)

Finally, factorise $a^2{b}^2(a^4-b^4)$ as follows:

$a^2{b}^2(a^4-b^4)=a^2{b}^2\left[\right(a^2{)}^2-\left(b^2{)}^2\right]=a^2{b}^2(a^2-b^2)(a^2+b^2)=a^2{b}^2(a^2+b^2)(a+b)(a-b)$

(using identity $a^2-b^2=(a+b)(a-b)$)

Since the common factor between the polynomials $a^{4}b-a{b}^4$,$a^4{b}^2-a^2{b}^4$ and $a^2{b}^2(a^4-b^4)$ is $ab(a-b)$.

Hence, the $HCF$ is $ab(a-b)$.