Question

Factorise:
$a^8-2a^4{b}^4+b^8$

• A. $\left[\right(a^2-b^2\left)\right(a+b\left)\right(a-b){]}^2$
• B. $\left[\right(a^2+b^2\left)\right(a+b\left)\right(a-b){]}^2$
• C. $\left[\right(a+b^2\left)\right(a+b\left)\right(a-b){]}^2$
• D. $\left[\right(a^2+b\left)\right(a+b\left)\right(a-b){]}^2$

Answer 1

The correct option is B

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

Using Identity: $(A-B{)}^2=A^2-2AB+B^2$
We can say, $A=a^4$ and $B=b^4$
$\Rightarrow a^8-2a^4{b}^4+b^8=(a^4-b^4{)}^2$
$=\left[\right(a^2+b^2\left)\right(a^2-b^2){]}^2$
$=\left[\right(a^2+b^2\left)\right(a+b\left)\right(a-b){]}^2$